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Why the real numbers need i
◆FRAME
Solve $x^2 + 1 = 0$.
Subtract 1 from both sides and you have $x^2 = -1$. Now take the square root, and the work stops. No real number squares to a negative. A positive times a positive is positive, a negative times a negative is also positive, and $0^2 = 0$. There is nothing left to try.
A first reading calls that equation unsolvable. It is worth being careful about what actually failed. The equation is fine. What ran out is the number system. Every number you have met so far sits somewhere on the real line, and no point on that line squares to $-1$.
So build the number that does. Call it i, and define it by one rule:
$i^2 = -1.$
That is the whole definition. It is not a value waiting to be computed, it is a rule to be used.
With i available, $x^2 = -1$ has two solutions, $x = i$ and $x = -i$. The second one works because $(-i)^2 = i^2 = -1$ as well.
Attach i to the reals and you get the COMPLEX NUMBERS: everything of the form $z = a + i b$ with a and b real. The real numbers have not been replaced. They are the case $b = 0$.
The equation was never unsolvable. Its solutions were not on the real line, and this chapter builds the place where they live.
Every real number is somewhere on the top rail, so this is the whole of the evidence. x squared plus 1 equals 0 asks for a landing the map never makes.
One new direction is the whole of the extension, and it is bought once. Every construction later in the chapter — the conjugate, the modulus, polar form, the roots of a quadratic — is read off this same plane rather than added to it.
Once $i^2 = -1$ is fixed, every power of i is already decided. And there turn out to be only four different answers.
Start multiplying and watch. $i^1 = i$. $i^2 = -1$, by definition. Then $i^3 = i^2 \times i = -i$, and $i^4 = i^2 \times i^2 = (-1)(-1) = 1$.
Now $i^5 = i^4 \times i = 1 \times i = i$, and the list starts over. The cycle has length four: $i$, then $-1$, then $-i$, then 1, then round again.
So for a power n, only one thing decides the answer — the remainder when n is divided by 4.
Remainder 1 gives $i$.
Remainder 2 gives $-1$.
Remainder 3 gives $-i$.
Remainder 0 gives 1.
The reason is $i^4 = 1$. Multiplying by 1 changes nothing, so every complete group of four powers can be dropped without touching the answer.
Divide the exponent by 4 and keep the remainder. That one step handles any power of i, however large.
Each lane collects exponents four apart, which is why the joined path is one descending run repeated rather than a new shape each time. Read any lane rightwards and the exponents grow while the value never moves, which is what it means to say the size of the exponent does not matter. The bottom lane is the one to pause on: those exponents divide evenly by 4, and the value there is 1 because a whole number of laps finished, not because anything was raised to the power zero.
Multiplying by i turns a point a quarter turn about the origin, and every power of i is one unit from the origin, so all of them land on this circle. Four quarter turns make one full turn, which is why the fourth power is back at 1 and the list starts again. That is the whole of the rule: only the remainder of the power on division by 4 decides which of the four values you get.
Worked example
Evaluate $i^{23}$ and $i^{-1}$
$23 = 4 \times 5 + 3$ divide the exponent by 4 — the remainder, 3, is all that decides the value of $i^{23}$.
$i^{23} = (i^4)^5 \times i^3 = 1^5 \times i^3 = -i$ the five complete cycles each contribute a factor of 1, so they vanish, leaving just $i^3$.
$i^{-1} = i^3/i^4 = i^3/1 = i^3 = -i$ a fraction with $i$ underneath is not a finished answer — use $i^4 = 1$ to move it upstairs.
$i^{-1} = -i = i^{23}$ the same value as $i^{23}$ — the cycle predicts that, since $-1$ and $23$ leave the same remainder, 3, on division by 4; a negative power is not a special case, so add 4 to the exponent until it turns positive and take the remainder as usual.
The rail runs leftwards for one reason only. i to the fourth is 1, so dividing by it as often as you like costs nothing, and that is the whole content of the one-line move the worked solution makes.
Twenty-three quarter turns from 1: five whole laps, then three quarters more.
Complex numbers add, subtract and multiply by ordinary algebra. Exactly one extra rule ever gets used, and it is $i^2 = -1$.
Take $z_1 = a + i b$ and $z_2 = c + i d$.
ADDITION and SUBTRACTION keep the two parts apart. Real goes with real, imaginary with imaginary: $z_1 \pm z_2 = (a \pm c) + i(b \pm d)$. Nothing mixes.
MULTIPLICATION starts the way any two brackets do. Expand $(a + i b)(c + i d)$ term by term and you get $a c + i a d + i b c + i^2 b d$.
The first three terms are what ordinary algebra would give. The fourth is where complex numbers part company with it. $i^2$ is $-1$, so $i^2 b d$ becomes $-b d$ — a REAL term, carrying a minus sign.
Collect the two kinds: $z_1 z_2 = (a c - b d) + i(a d + b c)$.
*The minus in the real part is not a misprint. It is the $i^2$, and losing it is the commonest error in the chapter.*
The sum is the far corner and the difference is the near side, so the two operations never needed separate pictures.
Worked example
Add, subtract, and multiply $3+2i$ and $1-4i$
$z_1 + z_2 = (3+1) + (2 + (-4))i = 4 - 2i$ the sum keeps the parts separate — real with real, imaginary with imaginary.
$z_1 - z_2 = (3-1) + (2-(-4))i = 2 + 6i$ the difference works the same way — watch that second subtraction: taking away $-4i$ ADDS.
$(3+2i)(1-4i) = 3 - 12i + 2i - 8i^2$ expand the product term by term, the way any two brackets would.
$-8i^2 = -8 \times (-1) = 8$ the rule that turns this term real — two minus signs cancel, and a term that looked imaginary has turned real.
$z_1 z_2 = (3+8) + (-12+2)i = 11 - 10i$ collect what is left, real with real and imaginary with imaginary — expand first, substitute $i^2=-1$ second, collect third, since working in that order keeps every sign change visible.
Three of the four terms stay in the column they started in. The fourth carries an i squared, so it turns real.
Asking when two complex numbers are equal has a sharper answer than it first looks.
$a + i b$ and $c + i d$ are equal exactly when $a = c$ AND $b = d$. Both conditions, and separately.
The reason is that the real and imaginary parts cannot trade with each other. No real number equals a nonzero imaginary one, so a shortfall in the real part can never be covered by the imaginary part.
What this gives you is a technique. One complex equation is secretly TWO real equations. If a problem hands you $x + i y = 5 - 3i$ with x and y real, read off $x = 5$ and $y = -3$ straight away. There is nothing to solve.
That is worth holding on to. It turns unknowns buried inside a complex expression into an ordinary pair of real equations, which you already know how to handle.
Split a complex equation into its real half and its imaginary half before doing anything else with it.
Each near-miss satisfies one of the two conditions and lands somewhere else entirely, which is what "both, and separately" means on the page.
The CONJUGATE of $z = a + i b$ is written $\overline{z}$, and it flips one sign only: $\overline{z} = a - i b$. The real part is untouched.
That looks like a small operation. Its whole use comes from one product.
Multiply a complex number by its own conjugate: $z \overline{z} = (a + i b)(a - i b)$.
Expand as usual and you get $a^2 - i a b + i a b - i^2 b^2$. The two middle terms cancel exactly — same size, opposite signs. And $-i^2 b^2$ is $+b^2$.
So $z \overline{z} = a^2 + b^2$. Real, every time. And never negative, being a sum of two squares.
That is the property to carry forward. Conjugation is the tool that turns a complex number REAL without throwing any of it away — both a and b are still there in the answer.
Multiplying by the conjugate is how you get i out of a place you do not want it.
The conjugate changes exactly one thing, the sign of the imaginary part, and on the plane that is a reflection in the real axis. The two triangles show what the word only is doing there: same base, same height, same distance from the origin. Multiply the two numbers and the imaginary parts cancel, leaving a squared plus b squared — which is that shared distance, squared.
▲CONCEPT
Dividing by a complex number needs no new rule. It needs a trick, and the last section is the trick.
The trouble with $1/(c + i d)$ is the i underneath. A denominator with i in it is not a finished answer, because you cannot read off the real and imaginary parts.
So make the denominator real. Multiply top and bottom by the denominator’s CONJUGATE, $c - i d$. Multiplying by $(c - i d)/(c - i d)$ is multiplying by 1, so the value does not change — only the way it is written.
That gives
$1/(c + i d) = (c - i d)/((c + i d)(c - i d)) = (c - i d)/(c^2 + d^2) \cdot$
The bottom is now the plain real number $c^2 + d^2$. Dividing by a real number is ordinary work: split the top and divide each part by it. So the answer in $a + i b$ form is
$c/(c^2 + d^2) - i (d/(c^2 + d^2)) \cdot$
The conjugate used is always the conjugate of the DENOMINATOR. The numerator does not get a say in choosing it.
Multiplying by the conjugate is chosen for one reason: it is the move that lands the bottom on the axis.
Worked example
Simplify $(3+2i)/(1-i)$ using the conjugate
$(3 + 2i)/(1 - i) \times (1 + i)/(1 + i)$ multiply top and bottom by the denominator’s conjugate — the conjugate to use is always the denominator’s.
$(1-i)(1+i) = 1 - i^2 = 1 + 1 = 2$ do the bottom first, because it is the easier half — real, as promised.
$(3+2i)(1+i) = 3+3i+2i+2i^2 = 1 + 5i$ the last term is $2 \times (-1) = -2$, so the real part is $3-2=1$ and the imaginary part is $3+2=5$.
$(3+2i)/(1-i) = (1+5i)/2 = 1/2 + (5/2) i$ put the two halves together, then split the fraction for strict $a + i b$ form — doing the denominator first paid off, since once it is known to be real, everything after it is a real division.
The two panels run in the worked example’s own order, and the halving at the end is the one step the written solution cannot show you.
Every complex number has a size, and the size is a real number.
The MODULUS of $z = a + i b$ is written $|z|$ and defined as $|z| = \sqrt{a^2 + b^2}$.
That formula should look familiar. It is Pythagoras. If a and b are two perpendicular distances, then $\sqrt{a^2 + b^2}$ is the straight-line distance across — and the next section makes that reading exact rather than suggestive.
The modulus and the conjugate are one idea seen twice. Recall $z \overline{z} = a^2 + b^2$, and compare it with $|z|^2 = a^2 + b^2$. The two right-hand sides are identical, so $|z|^2 = z \overline{z}$.
A value to fix it: for $z = 3 + 4i$,
$|z| = \sqrt{9 + 16} = \sqrt{25} = 5.$
Two things the modulus is not. It is not the real part. And it is never negative — a square root of a sum of squares cannot be.
The modulus throws away direction and keeps size. Different complex numbers can share one.
Same size, five different directions — the modulus keeps one and throws the other away.
A complex number carries two independent real numbers, a and b. Anything carrying two real numbers can be drawn as a point.
That is the ARGAND PLANE. Plot $z = a + i b$ at the point $(a, b)$. The horizontal axis carries the REAL part and is called the real axis; the vertical axis carries the IMAGINARY part and is called the imaginary axis.
So $3 + 4i$ sits three across and four up. A real number such as 5 sits on the horizontal axis at $(5, 0)$ — the reals have not gone anywhere, they are one line inside the plane.
Two earlier facts now have pictures.
Adding complex numbers is adding the points, coordinate by coordinate — the same rule as adding vectors. And the modulus $|z| = \sqrt{a^2 + b^2}$ is exactly the distance from the origin to the point, by Pythagoras on the two axes.
Complex numbers are not a longer number line. They are a plane, and the old line is the horizontal axis of it.
Both dots use the digits 3 and 4, and they are not the same point. That is the whole reason the order of the two parts is fixed rather than a matter of taste: reading a complex number backwards does not produce an untidy version of the right answer, it produces a different number altogether. The third dot is the other half of the picture. A real number still has a place here, on the horizontal axis and nowhere else, because its imaginary part is zero and there is nothing to go up by.
Adding two complex numbers adds the real parts and adds the imaginary parts, and that is exactly what happens when two arrows are added tip to tail. The sum lands on the far corner of the parallelogram. So complex addition is not a new rule to learn — it is the vector addition you already know, written in a different notation.
▲CONCEPT
A point in a plane can be located two ways: by its coordinates, or by how far away it is and in which direction. Complex numbers inherit both.
POLAR FORM uses the second. For any nonzero z, write $z = r (\cos(\theta) + i \sin(\theta))$, where r is the distance from the origin and $\theta$ is the angle measured from the positive real axis.
Both pieces have names. r is the modulus, $r = |z|$, which is nothing new. $\theta$ is the ARGUMENT of z, and that is the new part.
Check the form does what it claims. A point at distance r and angle $\theta$ has coordinates $r \cos(\theta)$ across and $r \sin(\theta)$ up. Write those as the real and imaginary parts and you get $z = r \cos(\theta) + i r \sin(\theta)$ — which is the form above, with r taken outside the bracket.
What polar form buys is a clean split. r is SIZE, $\theta$ is DIRECTION, and neither constrains the other.
Rectangular form is built for adding. Polar form is built for multiplying, where the sizes multiply and the angles add.
Rectangular form and polar form are two sets of directions to the same dot — nothing about the number changes.
Worked example
Convert $z=1+i$ to polar form
$r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}$ the modulus is direct — one of the two numbers polar form needs.
$\theta = \pi/4$ plot the point $z=1+i$ at $(1,1)$ to find the argument — one across, one up, at 45 degrees from the positive real axis: the first quadrant, on the diagonal.
$z = \sqrt{2} (\cos(\pi/4) + i \sin(\pi/4))$ substitute both r and theta into polar form.
$\sqrt{2}(1/\sqrt{2} + i/\sqrt{2}) = 1+i$ check it back — worth ten seconds, since landing the angle in the wrong quadrant is easy and multiplying back out catches it immediately; plot the point before naming the angle, because a formula for theta cannot see which quadrant you are in, but the picture can.
In practice, plot the point before taking the arctan. A calculator returns the first-quadrant turn whatever the signs were, and only the sketch says whether pi has to be added to it.
Polar form describes the same number with a different pair of numbers: how far it is from the origin, and which way it points. The distance is the modulus and the angle is the argument. Nothing has been added and nothing lost — the dashed drop recovers the coordinates you started from.
Now back to where the chapter started. For $a x^2 + b x + c = 0$ with $a \neq 0$, the roots are
$x = (-b \pm \sqrt{b^2 - 4 a c})/(2 a) \cdot$
The whole character of the answer sits under that square root, so the expression there gets a name: the DISCRIMINANT, $D = b^2 - 4 a c$.
There are three cases, and D alone decides which one you are in.
If $D > 0$, then $\sqrt{D}$ is a real number, and the $\pm$ pulls the two roots apart. Two distinct REAL roots.
If $D = 0$, then $\sqrt{D} = 0$, so the $\pm$ adds and subtracts nothing. Both roots are $-b/(2 a)$ — one repeated real root.
If $D < 0$, then $\sqrt{D}$ asks for the square root of a negative number. On the real line that is where the work used to stop. With i available it does not stop: write $\sqrt{D} = i \sqrt{-D}$, which is legitimate because $-D$ is positive, and carry on. Two COMPLEX roots.
The discriminant is the whole diagnosis and you can read it without solving anything. A negative D is not a failure. It is the case this chapter was built for.
All three parabolas are the same curve slid to a different height — shift the top one down and it becomes the middle one, then the bottom one. The discriminant is only a name for where that slide crosses zero.
Worked example
Solve $x^2+2x+5=0$
$D = b^2-4 a c = 2^2-4(1)(5) = 4-20 = -16$ check the discriminant before anything else — negative, so the roots are complex, known before a single root is computed.
$x = (-2 \pm \sqrt{-16})/2$ the quadratic formula, with the coefficients read off directly: $a=1$, $b=2$, $c=5$.
$\sqrt{-16} = \sqrt{16} \times \sqrt{-1} = 4i$ that split is safe here because only ONE of the two factors is negative — the chapter’s later misconception shows what goes wrong when both are.
$x = (-2 \pm 4i)/2 = -1 \pm 2i$ divide each term by 2 — the two roots, $-1+2i$ and $-1-2i$, share a real part and have opposite imaginary parts: a conjugate pair. Compute D first, every time — it tells you what kind of answer to expect, and getting a different kind means an arithmetic slip somewhere.
Left: the graph of x squared plus 2x plus 5, missing the real axis. Right: its two roots on the Argand plane.
Sometimes you need facts ABOUT the roots without needing the roots themselves. Two such facts come free.
For $a x^2 + b x + c = 0$ with roots $\alpha$ and $\beta$: $\alpha + \beta = -b/a$, and $\alpha \beta = c/a$.
Both fall straight out of the formula. Write the two roots as $\alpha = (-b + \sqrt{D})/(2 a)$ and $\beta = (-b - \sqrt{D})/(2 a)$.
ADD them. The two square-root terms have opposite signs and cancel, leaving $(-2 b)/(2 a)$, which is $-b/a$.
MULTIPLY them. The two brackets have the shape $(p + q)(p - q)$, which is $p^2 - q^2$. Here that gives $(b^2 - D)/(4 a^2)$, and since $D = b^2 - 4 a c$ the two $b^2$ terms cancel to leave $(4 a c)/(4 a^2) = c/a$.
Both times the square root disappeared. That is what makes these two particular combinations special, and why the difference of the roots, say, is not.
These hold whatever D is. Complex roots still have a real sum and a real product, which is exactly what makes a conjugate pair possible.
Neither shape needs the roots to be found first — m and d are read straight off a, b and c, and the two areas match whatever they turn out to be.
The axis of symmetry sits halfway between the roots; the vertical intercept is their product.
Worked example
Build a quadratic with roots 3 and $-5$
$S = 3+(-5) = -2$, $P = 3 \times (-5) = -15$ the sum-and-product route is quicker than multiplying out $(x-3)(x+5)$ directly, and it keeps working when the brackets would not be pleasant.
$x^2 - S x + P = 0$ the standard form for a quadratic whose roots have sum S and product P.
$x^2+2x-15=0$ substitute $S=-2$ and $P=-15$ and tidy — note the minus sign in front of S: with $a=1$, $b$ equals $-S$, not $S$, and dropping that minus is the standard slip.
$x^2+2x-15=(x-3)(x+5)$ check by factorising — the roots are 3 and $-5$, correct, and the route is unchanged when the roots are complex, where multiplying the brackets out directly would be a good deal more work.
The slipped equation is one this chapter has already drawn, for a different question, which is exactly why it looks right when it is not.
Look back over the chapter and there is only one new idea in it.
You added ONE number, i, with ONE rule, $i^2 = -1$. Everything since has been a consequence of that.
The ARITHMETIC is ordinary algebra plus one substitution. Add and subtract part by part; multiply by expanding the brackets and replacing $i^2$ with $-1$.
The CONJUGATE and the MODULUS are the two routes back to real numbers. The product $z \overline{z} = a^2 + b^2$ is what makes division possible, and $|z| = \sqrt{a^2 + b^2}$ measures size.
The ARGAND PLANE and POLAR FORM supply the picture. A complex number is a point; its modulus is the distance from the origin, and its argument is the direction.
And the QUADRATIC FORMULA says why any of this was needed. A negative discriminant is precisely where the real numbers run out of room, and $\sqrt{D}$ with $D < 0$ is the reason i had to be invented at all.
*One adjunction. When a step in this chapter stops making sense, take it back to $i^2 = -1$ and it will start again.*
The two arcs share a radius, which is the reason turning and measuring keep coming out of the same number.
The rule $\sqrt{a} \sqrt{b} = \sqrt{a b}$ is familiar and has never let you down, so it gets applied here too.
THE REALITY. The answer is $-1$, not 1. By definition $\sqrt{-1} = i$, and so
$\sqrt{-1} \times \sqrt{-1} = i \times i = i^2 = -1.$
Both cannot be right, and the definition wins. So the rule must carry a limit, and it does. $\sqrt{a} \sqrt{b} = \sqrt{a b}$ holds unless a and b are BOTH negative.
That exception is easy to state and easy to forget, because nothing in the working looks wrong when you break it. There is no warning sign — just an answer with the wrong sign in front of it.
The safe habit costs one extra step: convert to i FIRST, then multiply. So $\sqrt{-4} \times \sqrt{-9}$ is $(2i)(3i) = 6i^2 = -6$, and not $\sqrt{36} = 6$.
Two negatives under two separate roots is the one case where the product rule fails. Pull the i out before combining anything.
The two candidate answers are the two ends of one arc, so the argument is not a detail of the working — it is the whole difference between them.
✕MISCONCEPTION
THE TRAP. A quadratic with real coefficients has two complex roots, and they could be any two complex numbers that satisfy it — $2 + 3i$ and $7 - i$, say.
THE REALITY. They could not. If a real-coefficient quadratic has the complex root $p + i q$, then the other root is exactly $p - i q$. Complex roots of a real quadratic arrive in conjugate pairs, without exception.
The formula shows why. There is only one place a second root can come from: the $\pm$ in $x = (-b \pm \sqrt{D})/(2 a)$. Everything else — the $-b$, the $2 a$ — is identical for both roots.
So the two roots differ in one thing only, the sign in front of $\sqrt{D}$. When $D < 0$ that term is purely imaginary, so flipping its sign flips the imaginary part and leaves the real part alone. That is conjugation, exactly.
The sum and product check it in one line. $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$ are both REAL whenever a, b and c are. For $2 + 3i$ and $7 - i$ the sum is $9 + 2i$, which is not real — so that pair cannot be the roots of any real quadratic.
The condition is REAL coefficients. Allow the coefficients to be complex and the pairing rule goes away with them.
One glance settles a question the quadratic formula would take four lines to answer.
Exercise 4.1 — Arithmetic, conjugate, modulus, and polar form
practice Express $(3-2i)+(5+7i)$ in the form $a+i b$.
practice Find the multiplicative inverse of $2+3i$.
practice Find $|z|$ for $z = 4-3i$.
$5$
$7$
$1$
$25$
practice Simplify $(2+3i)/(1-2i)$ and express the result in the form $a+i b$.
practice Convert $z = -\sqrt{3} + i$ to polar form.
Answers
$8+5i$
$2/13 - (3/13) i$
$5$
$-4/5 + (7/5) i$
$z = 2 (\cos((5 \pi)/6) + i \sin((5 \pi)/6))$
Each wrong option is a real step done to 4 and 3 — the corner route, the difference, and the area of the square on the answer — so the one to pick is the one the swing lands on.
Miscellaneous — practice set
practice Solve $x^2 - 4x + 13 = 0$.
practice For $3x^2 + 5x - 2 = 0$ with roots $\alpha$ and $\beta$, find $\alpha + \beta$.
$-5/3$
$5/3$
$-2/3$
$2/3$
practice Evaluate $i^{37} + i^{-37}$.
practice A quadratic equation with real coefficients has one root $3-2i$. Find the equation.
Answers
$x = 2 \pm 3i$
$-5/3$
$0$
$x^2 - 6x + 13 = 0$
Having found the roots, both checks are lengths already on the page.
Four fresh exponents, four different answers, and the same three seconds of work for each.
Item 18 in this set asks which quadrant a number lies in, and this is the whole method. All four dots are built from the same two numbers, 5 and 3, with nothing changing but the signs, and each sign pattern lands in one region and no other. Worth noticing is what the answer does not require: no modulus, no argument, and no axes sketched in the margin. The two signs are already visible in the number as it is written.
The cycle of four values tells you what each power of i is; it does not tell you what a run of them adds to, and two items in these sets ask exactly that. Laid head to tail the four steps close a square, and a walk that ends where it began has contributed nothing. From there a long sum needs no arithmetic: a hundred consecutive powers is twenty-five of these closed walks, so the total is zero, and in any other run only the terms left over after the last complete four are worth evaluating.
Chapter-end problems — graded set
board-easy Evaluate $i^15$.
board-easy Which of these equals $i^38$?
$-1$
$1$
$i$
$-i$
board-easy Find $z_1+z_2$ for $z_1=4+3i$ and $z_2=-2+5i$.
board-easy Multiply $(2+3i)(1-2i)$.
board-easy Find real $x$ and $y$ such that $(x-2)+3i = 5+(y+1)i$.
board-easy Find $\overline{z}$ and $z \cdot \overline{z}$ for $z=5-3i$.
board-easy Find the modulus of $z=-8+6i$.
board-easy Solve $x^2+4x+13=0$ using the quadratic formula.
board-easy A quadratic has roots $\alpha=4$ and $\beta=-7$. Find the sum and product of the roots.
board-easy Divide $(4+2i)/(1+i)$.
board-standard Evaluate $i^2+i^4+i^6+i^8$.
board-standard Find $z_1-z_2$ for $z_1=2-6i$ and $z_2=-3+4i$.
board-standard Which of these equals $(3-i)(2+4i)$?
$10+10i$
$6+10i$
$10-10i$
$2+10i$
board-standard Which values of $x$ and $y$ satisfy $x + (y-2)i = -3+7i$?
$x=-3, y=9$
$x=3, y=9$
$x=-3, y=5$
$x=-3, y=7$
board-standard Find $z+\overline{z}$ and $z-\overline{z}$ for $z=6-9i$.
Answers
$-i$
$-1$
$2+8i$
$8-i$
$x=7, y=2$
$\overline{z}=5+3i$, $z \cdot \overline{z} = 34$
$10$
$x=-2+3i$ or $x=-2-3i$
sum $=-3$, product $=-28$
$3-i$
$0$
$5-10i$
$10+10i$
$x=-3, y=9$
$z+\overline{z}=12$, $z-\overline{z}=-18i$
Chapter-end problems — graded set (continued)
board-standard Divide $(5-3i)/(2+i)$.
board-standard Which of these equals $|3-4i|$?
$5$
$7$
$1$
$25$
board-standard In which quadrant of the Argand plane does $z=-4-7i$ lie?
board-standard Convert $z=\sqrt{3}+i$ to polar form.
board-standard Solve $x^2-6x+13=0$.
board-standard A quadratic equation has roots summing to $-7$ with product $10$. Find the equation.
board-standard A student writes $\sqrt{-4} \cdot \sqrt{-9} = \sqrt{36} = 6$. Is this correct? Find the correct value.
JEE One root of $x^2 - (3+i)x + (2+2i) = 0$ is $2$. Explain why the other root need not be $\overline{2}$.
JEE Convert $z=-1-i$ to polar form, with $\theta$ in $(-\pi,\pi]$.
JEE Find $z=(3+4i)/(1-2i)$, then find $|z|$ and $\overline{z}$.
JEE One root of $x^2+p x+q=0$, with $p$ and $q$ real, is $3-2i$. Find $p$ and $q$.
JEE Evaluate $i^37 + i^{-37}$.
JEE Describe the set of points $z=x+i y$ satisfying $|z-3|=|z-3i|$.
JEE Find $z=(1+i)^4$ by converting to polar form first.
JEE Which of these equals $|z_1 z_2|$ for $z_1=2+3i$ and $z_2=1-i$?
$\sqrt{26}$
$26$
$\sqrt{13}+\sqrt{2}$
$5$
Answers
$(7-11i)/5$
$5$
third quadrant
$z=2(\cos(\pi/6)+i \sin(\pi/6))$
$x=3+2i$ or $x=3-2i$
$x^2+7x+10=0$
incorrect — correct value is $-6$
not a conjugate pair — coefficients are complex, so the rule doesn’t apply
$z=\sqrt{2}(\cos(-3\pi/4)+i \sin(-3\pi/4))$
$z=-1+2i$, $|z|=\sqrt{5}$, $\overline{z}=-1-2i$
$p=-6, q=13$
$0$
the line $x=y$
$z=-4$
$\sqrt{26}$
Every root here is recomputed from its own question’s coefficients, so the picture is a check on the working rather than a copy of the answer key.
Every item in this set asks for the same move, which is to reach the answer through a property instead of by expanding. This is that move in its plainest form. The two factors measure 5 and root 2 from the origin, the product measures root 50, and that was readable off the picture before anyone multiplied 3 plus 4i by 1 minus i. The rule behind it holds for every pair of complex numbers, so it is worth carrying into the rest of the bank.
JEE-application problems — from the item bank
Evaluate $(1+i)^4$ $(1-i)^2$ by first simplifying each factor as a square, rather than expanding the full product term by term.
8i, since $(1+i)^2$ = 2i gives $(1+i)^4$ = $(2i)^2$ = -4, and $(1-i)^2$ = -2i, so the product is (-4)(-2i) = 8i.
-8i, from a sign slip while squaring (1-i), getting +2i instead of -2i, which flips the sign of the final product.
8, from multiplying only the real coefficients -4 and -2 and leaving the i off the result.
64, from treating $[(1+i)(1-i)]^6$ = $2^6$, as if the two unlike exponents 4 and 2 could be combined by sharing one base.
Check your answer
✓ 8i, since $(1+i)^2$ = 2i gives $(1+i)^4$ = $(2i)^2$ = -4, and $(1-i)^2$ = -2i, so the product is (-4)(-2i) = 8i. — (A) $(1+i)^2$ = 2i, so $(1+i)^4$ = $(2i)^2$ = -4. Separately, $(1-i)^2$ = -2i. Multiplying the two simplified factors, (-4)(-2i) = 8i.
-8i, from a sign slip while squaring (1-i), getting +2i instead of -2i, which flips the sign of the final product. — $(1-i)^2$ = 1 - 2i + $i^2$ = -2i, not +2i — the middle term keeps its minus sign. Using +2i by mistake flips the sign of the whole product to -8i.
8, from multiplying only the real coefficients -4 and -2 and leaving the i off the result. — -4 and -2i multiply to -4 times -2 times i = 8i — the i stays attached to the second factor throughout; it does not drop out.
64, from treating $[(1+i)(1-i)]^6$ = $2^6$, as if the two unlike exponents 4 and 2 could be combined by sharing one base. — $(1+i)^4$ and $(1-i)^2$ are different powers of different numbers — they must be simplified separately (to -4 and -2i) and then multiplied, not merged into one base first.
For z1 = 3+4i and z2 = 1-i, find |z1 z2| using |z1||z2|, without expanding the product z1 z2 first.
5 + $\sqrt{2}$, from adding the two moduli instead of multiplying them.
5 $\sqrt{2}$, since |z1| = 5 and |z2| = $\sqrt{2}$, so |z1 z2| = 5 $\sqrt{2}$.
5, from computing |z1+z2| (the modulus of the sum) instead of |z1 z2| (the modulus of the product).
10, from using $|\text{z2}|^2$ = 2 in place of |z2| = $\sqrt{2}$, skipping the square root.
Check your answer
5 + $\sqrt{2}$, from adding the two moduli instead of multiplying them. — |z1 z2| = |z1| times |z2|, not |z1| + |z2| — the modulus of a product multiplies, it does not add.
✓ 5 $\sqrt{2}$, since |z1| = 5 and |z2| = $\sqrt{2}$, so |z1 z2| = 5 $\sqrt{2}$. — (B) |z1| = $\sqrt{3^2+4^2}$ = 5 and |z2| = $\sqrt{1^2+1^2}$ = $\sqrt{2}$. Since |z1 z2| = |z1||z2|, the product’s modulus is 5 $\sqrt{2}$ — no need to expand z1 z2 first.
5, from computing |z1+z2| (the modulus of the sum) instead of |z1 z2| (the modulus of the product). — z1+z2 = 4+3i has modulus 5, but the question asks for |z1 z2|, the modulus of the PRODUCT, which uses |z1| times |z2| = 5 $\sqrt{2}$, not the modulus of the sum.
10, from using $|\text{z2}|^2$ = 2 in place of |z2| = $\sqrt{2}$, skipping the square root. — $|\text{z2}|^2$ = $1^2+(-1)^2$ = 2 is the modulus SQUARED. The modulus itself still needs the square root: |z2| = $\sqrt{2}$, giving |z1 z2| = 5 $\sqrt{2}$, not 5 times 2 = 10.
For z1 = 1+i and z2 = $\sqrt{3}-i$, find |z1/z2| using |z1|/|z2|, without dividing z1 by z2 first.
$\sqrt{2}$, from inverting the ratio — dividing |z2| by |z1| instead of |z1| by |z2|.
2 $\sqrt{2}$, from multiplying the two moduli instead of dividing them.
$\sqrt{2}/2$, since |z1| = $\sqrt{2}$ and |z2| = 2, so |z1/z2| = $\sqrt{2}/2.$
$\sqrt{2}/4$, from using $|\text{z2}|^2$ = 4 in the denominator instead of |z2| = 2.
Check your answer
$\sqrt{2}$, from inverting the ratio — dividing |z2| by |z1| instead of |z1| by |z2|. — |z1/z2| = |z1|/|z2|, with |z1| in the numerator — z1 is being divided BY z2, so |z1| = $\sqrt{2}$ belongs on top, not |z2| = 2.
2 $\sqrt{2}$, from multiplying the two moduli instead of dividing them. — The question asks for |z1/z2|, a quotient — |z1| and |z2| must be divided, not multiplied.
✓ $\sqrt{2}/2$, since |z1| = $\sqrt{2}$ and |z2| = 2, so |z1/z2| = $\sqrt{2}/2.$ — (C) |z1| = $\sqrt{1^2+1^2}$ = $\sqrt{2}$ and |z2| = $\sqrt{(\sqrt{3})^2+(-1)^2}$ = $\sqrt{4}$ = 2. Since |z1/z2| = |z1|/|z2|, the quotient’s modulus is $\sqrt{2}/2.$
$\sqrt{2}/4$, from using $|\text{z2}|^2$ = 4 in the denominator instead of |z2| = 2. — $|\text{z2}|^2$ = $(\sqrt{3})^2+(-1)^2$ = 4 is the modulus SQUARED. The modulus itself is |z2| = $\sqrt{4}$ = 2, which is what belongs in the denominator, giving $\sqrt{2}/2$, not $\sqrt{2}/4.$
Simplify (1+i)/(1-i) + (1-i)/(1+i).
2i, from computing the second term as if it equalled the first term, i + i instead of i + (-i).
-2i, from a sign error that makes both terms come out as -i instead of one being +i and the other -i.
2, from dropping the imaginary unit off each simplified term, treating i and -i as if they were 1 and 1.
0, since (1+i)/(1-i) = i and (1-i)/(1+i) = -i, and i + (-i) = 0.
Check your answer
2i, from computing the second term as if it equalled the first term, i + i instead of i + (-i). — (1+i)/(1-i) simplifies to i, but (1-i)/(1+i) is a DIFFERENT fraction (numerator and denominator swapped) and simplifies to -i, not i again.
-2i, from a sign error that makes both terms come out as -i instead of one being +i and the other -i. — Multiplying (1+i)/(1-i) by (1+i)/(1+i) gives $(1+i)^2/2$ = 2i/2 = i, a POSITIVE i — only the second term, (1-i)/(1+i), simplifies to -i.
2, from dropping the imaginary unit off each simplified term, treating i and -i as if they were 1 and 1. — The two simplified terms are i and -i, not 1 and 1 — the imaginary unit is part of the value, not a label to drop before adding.
✓ 0, since (1+i)/(1-i) = i and (1-i)/(1+i) = -i, and i + (-i) = 0. — (D) (1+i)/(1-i) = $(1+i)^2/[(1-i)(1+i)]$ = 2i/2 = i. Similarly (1-i)/(1+i) = -2i/2 = -i. Adding, i + (-i) = 0.
z1 = 2(cos30° + i sin30°) and z2 = 3(cos60° + i sin60°). Find z1 z2 in polar form, using the polar multiplication rule (moduli multiply, arguments add) rather than converting to rectangular form first.
6(cos90° + i sin90°) = 6i, since the moduli multiply (2×3=6) and the arguments add (30°+60°=90°).
5(cos90° + i sin90°) = 5i, from adding the two moduli (2+3=5) instead of multiplying them.
6(cos0° + i sin0°) = 6, from multiplying the two arguments (30°×60°=1800°, which reduces to 0°) instead of adding them.
6(cos30° + i sin30°) = 3 $\sqrt{3}$ + 3i, from using only z1’s argument and forgetting to add z2’s argument.
Check your answer
✓ 6(cos90° + i sin90°) = 6i, since the moduli multiply (2×3=6) and the arguments add (30°+60°=90°). — (A) For a product in polar form, moduli multiply and arguments add — r = 2×3 = 6, θ = 30°+60° = 90°. So z1 z2 = 6(cos90°+i sin90°) = 6(0+i) = 6i.
5(cos90° + i sin90°) = 5i, from adding the two moduli (2+3=5) instead of multiplying them. — Multiplying two complex numbers in polar form multiplies their moduli (2 times 3 = 6); it does not add them (2+3=5 is not the modulus of the product).
6(cos0° + i sin0°) = 6, from multiplying the two arguments (30°×60°=1800°, which reduces to 0°) instead of adding them. — The arguments of a product ADD (30°+60°=90°); they do not multiply — multiplying 30° by 60° has no meaning here and happens to reduce to a misleading 0°.
6(cos30° + i sin30°) = 3 $\sqrt{3}$ + 3i, from using only z1’s argument and forgetting to add z2’s argument. — Both arguments contribute — the product’s argument is the SUM of both, 30°+60°=90°, not z1’s argument alone.
Solve $x^2$ - 2ix - 1 = 0 for x, using the quadratic formula with a=1, b=-2i, c=-1.
x = -i (a repeated root), from a sign error using -b = -2i directly instead of flipping the sign of b=-2i.
x = i (a repeated root), since D = $(-2i)^2$ - 4(1)(-1) = -4+4 = 0, so x = (2i ± 0)/2 = i.
x = i + $\sqrt{2}$ or x = i - $\sqrt{2}$, two distinct roots, from computing $(-2i)^2$ as +4 instead of -4.
x = 0 or x = 2i, from omitting the -4ac term entirely and using D = $(-2i)^2$ = -4 alone.
Check your answer
x = -i (a repeated root), from a sign error using -b = -2i directly instead of flipping the sign of b=-2i. — b = -2i, so -b = -(-2i) = 2i, not -2i — the sign must flip when moving from b to -b in the formula.
✓ x = i (a repeated root), since D = $(-2i)^2$ - 4(1)(-1) = -4+4 = 0, so x = (2i ± 0)/2 = i. — (B) D = $(-2i)^2$ - 4(1)(-1) = -4 + 4 = 0. x = [-(-2i) ± $\sqrt{0}]/2$ = (2i ± 0)/2 = i — a repeated root, since D = 0.
x = i + $\sqrt{2}$ or x = i - $\sqrt{2}$, two distinct roots, from computing $(-2i)^2$ as +4 instead of -4. — $(-2i)^2$ = $4i^2$ = 4(-1) = -4, not +4 — squaring -2i still picks up $i^2$ = -1, it does not become a positive real number.
x = 0 or x = 2i, from omitting the -4ac term entirely and using D = $(-2i)^2$ = -4 alone. — The discriminant is $b^2$ - 4ac, both terms — dropping -4ac = -4(1)(-1) = 4 leaves D = -4 instead of the correct D = -4+4 = 0.
A quadratic equation with real coefficients has one root 3+2i. Using the sum and product of its roots, find its discriminant D.
88, from using D = $(\sum)^2$ + 4(product) instead of $(\sum)^2$ - 4(product).
16, from computing the product of the roots as $3^2$ - $2^2$ = 5 instead of $3^2$ + $2^2$ = 13.
-16, since sum = 6 and product = 13, so D = $6^2$ - 4(13) = 36 - 52 = -16.
36, from using only $(\sum)^2$ as the discriminant and forgetting to subtract 4 times the product.
Check your answer
88, from using D = $(\sum)^2$ + 4(product) instead of $(\sum)^2$ - 4(product). — D = $(\sum)^2$ - 4(product), with a MINUS sign — adding instead gives 36+52=88, the wrong value and the wrong sign for a complex-root case.
16, from computing the product of the roots as $3^2$ - $2^2$ = 5 instead of $3^2$ + $2^2$ = 13. — The other root is the conjugate 3-2i, and (3+2i)(3-2i) = $3^2+2^2$ = 13 (the cross terms cancel, leaving a PLUS), not $3^2-2^2=5.$
✓ -16, since sum = 6 and product = 13, so D = $6^2$ - 4(13) = 36 - 52 = -16. — (C) The other root is the conjugate 3-2i, so sum = (3+2i)+(3-2i) = 6 and product = (3+2i)(3-2i) = 9+4 = 13. D = $\sum^2$ - 4(product) = 36 - 52 = -16.
36, from using only $(\sum)^2$ as the discriminant and forgetting to subtract 4 times the product. — The discriminant needs both terms — D = $(\sum)^2$ - 4(product) = 36 - 52 = -16, not just 36 with the product term left out.
A quadratic $2x^2$ + bx + c = 0 has real coefficients and one root 1+i. Find b and c.
b=-2, c=2, from ignoring the leading coefficient a=2 and treating the equation as if it were monic.
b=4, c=4, from dropping the minus sign in sum = -b/a, giving b the wrong sign.
b=-4, c=0, from computing the product (1+i)(1-i) as 1-1=0 instead of 1+1=2.
b=-4, c=4, since the other root is 1-i, sum = 2 = -b/2 gives b=-4, and product = 2 = c/2 gives c=4.
Check your answer
b=-2, c=2, from ignoring the leading coefficient a=2 and treating the equation as if it were monic. — With a=2, sum = -b/a = 2 gives b = -4 (not -2), and product = c/a = 2 gives c = 4 (not 2) — the leading coefficient a=2 must be carried through both formulas.
b=4, c=4, from dropping the minus sign in sum = -b/a, giving b the wrong sign. — sum = -b/a means b = -a(sum) = -2(2) = -4 — dropping the minus sign gives b=4, the wrong sign.
b=-4, c=0, from computing the product (1+i)(1-i) as 1-1=0 instead of 1+1=2. — (1+i)(1-i) = 1 - $i^2$ = 1-(-1) = 1+1 = 2, not 1-1=0 — the $i^2$ term becomes +1 once $i^2=-1$ is substituted, so the product is a sum of squares, not a difference.
✓ b=-4, c=4, since the other root is 1-i, sum = 2 = -b/2 gives b=-4, and product = 2 = c/2 gives c=4. — (D) With real coefficients, the other root is the conjugate 1-i. Sum = (1+i)+(1-i) = 2 = -b/2, so b=-4. Product = (1+i)(1-i) = 1+1 = 2 = c/2, so c=4.
A quadratic with real coefficients has sum of roots 4 and product of roots 13. Find the roots.
2+3i and 2-3i, since $x^2-4x+13=0$ gives D=16-52=-36, so x=(4±6i)/2=2±3i.
4+6i and 4-6i, from forgetting to divide the whole numerator (4±6i) by 2a=2.
5 and -1, from treating $\sqrt{-36}$ as the real number -6 instead of the imaginary number 6i.
2+6i and 2-6i, from dividing only the real part of the numerator by 2, and leaving the imaginary part 6i undivided.
Check your answer
✓ 2+3i and 2-3i, since $x^2-4x+13=0$ gives D=16-52=-36, so x=(4±6i)/2=2±3i. — (A) The quadratic is $x^2$ - (sum)x + (product) = $x^2-4x+13=0.$ D = 16-52 = -36, so x = [4 ± $\sqrt{-36}]/2$ = (4±6i)/2 = 2±3i.
4+6i and 4-6i, from forgetting to divide the whole numerator (4±6i) by 2a=2. — The full numerator 4±6i must be divided by 2a=2, giving 2±3i — leaving it undivided gives roots twice the correct size.
5 and -1, from treating $\sqrt{-36}$ as the real number -6 instead of the imaginary number 6i. — D=-36 is negative, so $\sqrt{-36}$ = 6i (imaginary), not -6 (real) — treating it as real produces two real roots when the equation actually has none.
2+6i and 2-6i, from dividing only the real part of the numerator by 2, and leaving the imaginary part 6i undivided. — Both the real part (4) and the imaginary part (6i) of the numerator must be divided by 2a=2 — dividing only the 4 leaves the imaginary part too large.
Evaluate the sum $i^1$ + $i^2$ + $i^3$ + ... + $i^100.$
1, from treating the 100th term as a leftover extra term $i^0=1$ added after 25 complete cycles, instead of the cycles alone summing to 0.
0, since the sum splits into 25 complete four-term cycles (i + (-1) + (-i) + 1 = 0 each), and 25 times 0 is 0.
50i - 50, from treating the cycle length as 2 instead of 4, alternating i, -1, i, -1, ...
100, from assuming every one of the 100 terms contributes +1 to the sum.
Check your answer
1, from treating the 100th term as a leftover extra term $i^0=1$ added after 25 complete cycles, instead of the cycles alone summing to 0. — 100 is exactly divisible by 4 (25 complete cycles), so there is no leftover term to add — each full cycle i+(-1)+(-i)+1 already sums to 0.
✓ 0, since the sum splits into 25 complete four-term cycles (i + (-1) + (-i) + 1 = 0 each), and 25 times 0 is 0. — (B) The sum has 100 terms, forming exactly 25 complete four-term cycles (i^1 through $i^4$, $i^5$ through $i^8$, and so on). Each cycle sums to i + (-1) + (-i) + 1 = 0, so the total is 25 × 0 = 0.
50i - 50, from treating the cycle length as 2 instead of 4, alternating i, -1, i, -1, ... — The powers of i repeat every 4 terms (i, -1, -i, 1), not every 2 — $i^1=i$ and $i^3=-i$ are different values, so a length-2 cycle is not consistent with the actual pattern.
100, from assuming every one of the 100 terms contributes +1 to the sum. — Only $i^4$, $i^8$, $i^12$, and so on equal 1 — most terms in the sum are i, -1, or -i, not 1, so not every term contributes +1.
The item this sits above asks how large the modulus of z plus 3i can be when the modulus of z is 2, and its three wrong options are the three ways that question gets answered without a picture: the smallest value reported as the largest, the two sizes combined as though they were the legs of a right triangle, and the shift forgotten altogether. Each of those is a guess about a distance that the drawing simply measures. The same reading settles every later item in this bank that fixes one modulus and asks what another one can be.
This set’s reciprocal item says in its own stem to use the identity rather than rationalise the fraction from scratch, and the picture is what that instruction buys. Rationalising is a calculation whose answer arrives with no sense of where it belongs; the identity is two moves whose result can be placed before any arithmetic is done, which also makes a wrong answer visible as a point in the wrong region. The chapter’s earlier division figures show a fraction being cleared, which is a different question from where the answer lands.
JEE-application problems — continued
Solve $x^2$ - 2x + 5 = 0, then find the modulus of either root.
2√5, from the undivided root.
3, adding real and imaginary parts.
√5, from the roots 1±2i.
5, the modulus squared, not the modulus.
Check your answer
2√5, from the undivided root. — The root itself is 1+2i, not 2+4i — the numerator (2±4i) must be divided by 2a=2 BEFORE its modulus is taken, not after.
3, adding real and imaginary parts. — The modulus of 1+2i is $\sqrt{1^2+2^2}$ = $\sqrt{5}$, a distance computed with squares and a square root — not the plain sum of the parts, 1+2=3.
✓ √5, from the roots 1±2i. — (C) D = 4-20 = -16, so x = (2 ± 4i)/2 = 1 ± 2i. Taking either root, |1+2i| = $\sqrt{1^2+2^2}$ = $\sqrt{5}$.
5, the modulus squared, not the modulus. — $|1+2i|^2$ = $1^2+2^2$ = 5 is the modulus SQUARED. The modulus itself still needs the square root: |1+2i| = $\sqrt{5}$.
Evaluate $[(1+i)/(1-i)]^10$, using the fact that (1+i)/(1-i) simplifies to a power of i first.
1, mis-simplifying without the conjugate.
-i, misreading the exponent remainder.
i, misreading the exponent remainder.
-1, from (1+i)/(1-i)=i and $i^10=-1.$
Check your answer
1, mis-simplifying without the conjugate. — (1+i)/(1-i) does not cancel to 1 — multiplying top and bottom by the conjugate 1+i gives $(1+i)^2/2$ = 2i/2 = i, not 1.
-i, misreading the exponent remainder. — 10 divided by 4 leaves remainder 2 (4×2=8, 10-8=2), not remainder 3 — $i^10$ = $i^2$, not $i^3.$
i, misreading the exponent remainder. — 10 divided by 4 leaves remainder 2, not remainder 1 — $i^10$ = $i^2$ = -1, not $i^1$ = i.
✓ -1, from (1+i)/(1-i)=i and $i^10=-1.$ — (D) Multiplying by the conjugate, (1+i)/(1-i) = $(1+i)^2/[(1-i)(1+i)]$ = 2i/2 = i. Then $i^10$ = $i^{4×2+2}$ = $i^2$ = -1, since 10 divided by 4 leaves remainder 2.
If |z| = 2, what is the maximum possible value of |z + 3i| as z varies over all points satisfying |z| = 2?
5, since |z+3i| ≤ |z| + |3i| = 2+3 = 5, with equality when z sits in the same direction as 3i (z=2i).
1, from computing the MINIMUM value, ||z| - |3i|| = |2-3| = 1, instead of the maximum.
$\sqrt{13}$, from wrongly combining the two moduli with the Pythagorean formula, $\sqrt{2^2+3^2}$, instead of adding them directly.
2, from reporting |z| alone and forgetting to account for the +3i shift at all.
Check your answer
✓ 5, since |z+3i| ≤ |z| + |3i| = 2+3 = 5, with equality when z sits in the same direction as 3i (z=2i). — (A) By the triangle inequality, |z+3i| ≤ |z|+|3i| = 2+3 = 5, and this maximum is reached exactly when z and 3i point in the same direction, i.e. z=2i.
1, from computing the MINIMUM value, ||z| - |3i|| = |2-3| = 1, instead of the maximum. — 1 is the SMALLEST possible value of |z+3i| (when z points opposite to 3i) — the question asks for the maximum, which is |z|+|3i|=5, when z and 3i point the same way.
$\sqrt{13}$, from wrongly combining the two moduli with the Pythagorean formula, $\sqrt{2^2+3^2}$, instead of adding them directly. — |z+3i| is bounded by the direct sum |z|+|3i|, following from z and 3i being added as points, not combined as legs of a right triangle: there is no right angle forced here.
2, from reporting |z| alone and forgetting to account for the +3i shift at all. — Adding 3i to z shifts every point by 3 units — the maximum distance from the origin after that shift is |z|+|3i|=5, not |z|=2 alone.
If (x+iy)(1+i) = 3+5i for real numbers x and y, find x and y.
x=4, y=-1, from swapping which combination is the real part and which is the imaginary part — setting x+y=3 and x-y=5.
x=4, y=1, since (x+iy)(1+i) = (x-y) + i(x+y), giving x-y=3 and x+y=5.
x=3, y=5, from reading the real and imaginary parts of 3+5i directly off as x and y, without expanding the product at all.
x=1, y=4, from a sign error that flips the real-part equation to y-x=3 instead of x-y=3.
Check your answer
x=4, y=-1, from swapping which combination is the real part and which is the imaginary part — setting x+y=3 and x-y=5. — Expanding (x+iy)(1+i) gives (x-y) + i(x+y) — the real part is x-y, and the imaginary part is x+y, not the other way round.
✓ x=4, y=1, since (x+iy)(1+i) = (x-y) + i(x+y), giving x-y=3 and x+y=5. — (B) (x+iy)(1+i) = $x+\text{xi}+\text{iy}+i^2y$ = (x-y)+i(x+y). Matching to 3+5i, x-y=3 and x+y=5. Adding these, 2x=8, so x=4; then y=1.
x=3, y=5, from reading the real and imaginary parts of 3+5i directly off as x and y, without expanding the product at all. — x and y are not the real and imaginary parts of 3+5i directly — the LEFT side must first be expanded, (x-y)+i(x+y), before matching real parts and imaginary parts.
x=1, y=4, from a sign error that flips the real-part equation to y-x=3 instead of x-y=3. — Expanding correctly gives real part x-y (not y-x) — flipping that sign changes which of x and y comes out larger.
Evaluate $\sqrt{-3}$ × $\sqrt{-12}$, then add $i^100$ to the result.
7, from an invalid square-root product rule.
-6+i, misreading the exponent remainder.
-5, the correct value.
-7, misreading the exponent remainder.
Check your answer
7, from an invalid square-root product rule. — $\sqrt{-3}$ and $\sqrt{-12}$ are both negative under the root, so $\sqrt{a}\sqrt{b}=\sqrt{\text{ab}}$ does not apply — the correct route is $\sqrt{-3}=i$ $\sqrt{3}$, $\sqrt{-12}=2i$ $\sqrt{3}$, giving a product of -6, not 6.
-6+i, misreading the exponent remainder. — 100 divided by 4 leaves remainder 0 exactly (4×25=100) — $i^100=i^4=1$, not $i^1=i$.
✓ -5, the correct value. — (C) $\sqrt{-3}=i$ $\sqrt{3}$ and $\sqrt{-12}=2i$ $\sqrt{3}$, so their product is $2i^2(3)=-6.$ Since 100 is exactly divisible by 4, $i^100=1.$ Adding, -6+1=-5.
-7, misreading the exponent remainder. — 100 divided by 4 leaves remainder 0, not remainder 2 — $i^100=1$, not $i^2=-1.$
Find $|i^37$ + $i^38$ + $i^39$ + $i^40|$.
4, from adding the individual magnitudes |i|+|-1|+|-i|+|1| = 1+1+1+1, instead of adding the terms first.
2, from computing $i^40$ with a remainder of 2 instead of 0, giving -1 instead of 1, so the sum becomes i-1-i-1=-2.
$\sqrt{2}$, from computing $i^37$ with a remainder of 0 instead of 1, giving 1 instead of i, so the sum becomes 1-1-i+1=1-i.
0, since $i^37=i$, $i^38=-1$, $i^39=-i$, $i^40=1$ (remainders 1,2,3,0), and i-1-i+1=0.
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4, from adding the individual magnitudes |i|+|-1|+|-i|+|1| = 1+1+1+1, instead of adding the terms first. — The terms must be ADDED first (they can cancel, since i and -i, and -1 and 1, cancel each other), and only THEN should the modulus of the total be taken — adding the magnitudes first throws away that cancellation.
2, from computing $i^40$ with a remainder of 2 instead of 0, giving -1 instead of 1, so the sum becomes i-1-i-1=-2. — 40 divided by 4 leaves remainder 0 exactly — $i^40=i^4=1$, not $i^2=-1.$
$\sqrt{2}$, from computing $i^37$ with a remainder of 0 instead of 1, giving 1 instead of i, so the sum becomes 1-1-i+1=1-i. — 37 divided by 4 leaves remainder 1 (4×9=36, 37-36=1) — $i^37=i^1=i$, not $i^4=1.$
✓ 0, since $i^37=i$, $i^38=-1$, $i^39=-i$, $i^40=1$ (remainders 1,2,3,0), and i-1-i+1=0. — (D) 37, 38, 39, 40 leave remainders 1, 2, 3, 0 when divided by 4, so $i^37=i$, $i^38=-1$, $i^39=-i$, $i^40=1.$ Summing, i + (-1) + (-i) + 1 = 0, and |0|=0.
For z1 = 1+2i and z2 = 2-i, find |z1 z2| using |z1||z2|, without expanding the product first.
5, since |z1| = $\sqrt{5}$ and |z2| = $\sqrt{5}$, so |z1 z2| = $\sqrt{5}$ × $\sqrt{5}$ = 5.
2 $\sqrt{5}$, from adding the two moduli, $\sqrt{5}+\sqrt{5}$, instead of multiplying them.
$\sqrt{13}$, from expanding z1 z2 but dropping the $-2i^2$ term (treating it as 0), giving 2+3i instead of the correct 4+3i.
25, from using the moduli SQUARED, $|\text{z1}|^2$ × $|\text{z2}|^2$ = 5×5, instead of the moduli themselves.
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✓ 5, since |z1| = $\sqrt{5}$ and |z2| = $\sqrt{5}$, so |z1 z2| = $\sqrt{5}$ × $\sqrt{5}$ = 5. — (A) |z1| = $\sqrt{1^2+2^2}$ = $\sqrt{5}$ and |z2| = $\sqrt{2^2+(-1)^2}$ = $\sqrt{5}$. Since |z1 z2| = |z1||z2|, the product’s modulus is $\sqrt{5}×\sqrt{5}$ = 5.
2 $\sqrt{5}$, from adding the two moduli, $\sqrt{5}+\sqrt{5}$, instead of multiplying them. — |z1 z2| = |z1| times |z2|, not |z1|+|z2| — the modulus of a product multiplies, it does not add.
$\sqrt{13}$, from expanding z1 z2 but dropping the $-2i^2$ term (treating it as 0), giving 2+3i instead of the correct 4+3i. — z1 z2 = (1+2i)(2-i) = $2-i+4i-2i^2$ = $2+3i-2i^2.$ Since $i^2=-1$, $-2i^2=+2$, giving 4+3i, not 2+3i — dropping that term loses the +2 contribution to the real part.
25, from using the moduli SQUARED, $|\text{z1}|^2$ × $|\text{z2}|^2$ = 5×5, instead of the moduli themselves. — $|\text{z1}|^2$ and $|\text{z2}|^2$ are the moduli SQUARED (5 and 5). The moduli themselves are $\sqrt{5}$ and $\sqrt{5}$, whose product is 5, not 25.
If (a+bi)/(1+i) = 2-3i for real numbers a and b, find a and b.
a=2, b=-1, from dropping the $-3i^2$ term in the expansion, treating it as 0 instead of +3.
a=5, b=1, from a sign error that drops the negative sign on the imaginary part of the result.
a=-1, b=-5, from multiplying (2-3i) by 1-i instead of by 1+i.
Check your answer
a=2, b=-1, from dropping the $-3i^2$ term in the expansion, treating it as 0 instead of +3. — $-3i^2$ becomes +3 once $i^2=-1$ is substituted, not 0 — dropping that term loses the +3 contribution to the real part, giving 2 instead of 5.
✓ a=5, b=-1, since a+bi = (2-3i)(1+i) = $2+2i-3i-3i^2$ = 2-i+3 = 5-i. — (B) Multiplying both sides by 1+i, a+bi = (2-3i)(1+i) = $2+2i-3i-3i^2$ = 2-i+3 = 5-i. So a=5 and b=-1.
a=5, b=1, from a sign error that drops the negative sign on the imaginary part of the result. — (2-3i)(1+i) = 2-i+3 = 5-i — the imaginary part is -1, not +1; there is no sign to drop here.
a=-1, b=-5, from multiplying (2-3i) by 1-i instead of by 1+i. — To isolate a+bi, both sides must be multiplied by 1+i (the SAME factor that is dividing, to clear the denominator directly) — multiplying by its conjugate 1-i instead changes the equation.
Using the identity 1/z = overline(z) / $|z|^2$ (rather than rationalizing from scratch), find 1/z for z = 1+i.
(1+i)/2, from using z itself in the numerator instead of its conjugate overline(z).
$(1-i)/\sqrt{2}$, from using |z| = $\sqrt{2}$ in the denominator instead of $|z|^2$ = 2.
(1-i)/2, since overline(z) = 1-i and $|z|^2$ = $1^2+1^2$ = 2, so 1/z = (1-i)/2.
undefined, from computing $|z|^2$ as $1^2-1^2$ = 0 instead of $1^2+1^2$ = 2.
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(1+i)/2, from using z itself in the numerator instead of its conjugate overline(z). — The identity uses overline(z) (the CONJUGATE), 1-i, in the numerator — using z itself, 1+i, does not cancel the imaginary part correctly when checking z times (1/z) = 1.
$(1-i)/\sqrt{2}$, from using |z| = $\sqrt{2}$ in the denominator instead of $|z|^2$ = 2. — The identity’s denominator is $|z|^2$ = 2, not |z| = $\sqrt{2}$ — using |z| alone leaves an extra factor of $\sqrt{2}$ in the result.
✓ (1-i)/2, since overline(z) = 1-i and $|z|^2$ = $1^2+1^2$ = 2, so 1/z = (1-i)/2. — (C) overline(z) = 1-i and $|z|^2$ = $1^2+1^2$ = 2. By the identity 1/z = overline(z) / $|z|^2$, 1/z = (1-i)/2 — no need to rationalize (1)/(1+i) from scratch.
undefined, from computing $|z|^2$ as $1^2-1^2$ = 0 instead of $1^2+1^2$ = 2. — $|z|^2$ = $a^2+b^2$ (a sum of squares, from z times overline(z)) = $1^2+1^2$ = 2, not $1^2-1^2$ = 0 — a and b are added as squares, never subtracted, and z=1+i is nonzero, so 1/z is well-defined.
Solve $x^2$ - (3+i)x + (2+i) = 0. (Note — the coefficients here are complex, not real; the real-coefficient conjugate-pair rule does not apply.)
x=2+i or x=2-i, wrongly assuming the roots form a conjugate pair since the discriminant contains i — but the conjugate-pair rule needs REAL coefficients, which this equation does not have.
x=2 or x=1+i, from a sign error computing $\sqrt{2i}$ as 1-i instead of 1+i.
x=4+2i or x=2, from forgetting to divide the numerator (3+i)±(1+i) by 2a=2.
x=1 or x=2+i, since $D=(3+i)^2-4(2+i)=2i$, $\sqrt{2i}=1+i$ (as (1+i)^2=2i), so x=[(3+i)±(1+i)]/2 gives x=2+i or x=1.
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x=2+i or x=2-i, wrongly assuming the roots form a conjugate pair since the discriminant contains i — but the conjugate-pair rule needs REAL coefficients, which this equation does not have. — The conjugate-pair rule (a complex root’s partner is always its conjugate) only holds when the coefficients are REAL. Here b=-(3+i) and c=2+i are themselves complex, so there is no such guarantee — the actual roots, 1 and 2+i, are not conjugates of each other.
x=2 or x=1+i, from a sign error computing $\sqrt{2i}$ as 1-i instead of 1+i. — $(1+i)^2$ = $1+2i+i^2$ = 2i, so $\sqrt{2i}$ = 1+i, not 1-i — using 1-i does not square back to 2i.
x=4+2i or x=2, from forgetting to divide the numerator (3+i)±(1+i) by 2a=2. — The numerator (3+i)±(1+i) must be divided by 2a=2 to finish the formula — skipping that division leaves both roots twice their correct size.
✓ x=1 or x=2+i, since $D=(3+i)^2-4(2+i)=2i$, $\sqrt{2i}=1+i$ (as (1+i)^2=2i), so x=[(3+i)±(1+i)]/2 gives x=2+i or x=1. — (D) D = $(3+i)^2$ - 4(2+i) = (9+6i-1) - (8+4i) = 2i. Since $(1+i)^2=2i$, $\sqrt{2i}=1+i$. x = [(3+i)±(1+i)]/2 gives x=(4+2i)/2=2+i or x=2/2=1.
The arms grow and turn by the same amount every time, which is the one thing expanding the brackets would never let you see.
Read as a distance rather than as an expression to square out, the condition stops being algebra at all.