Complex Numbers and Quadratic Equations

How many complex numbers satisfy $z^2=\bar{z}$?

It looks like a quadratic, so the expected answer is two. There are four.

Take the modulus of both sides: $|z{|}^2=|\bar{z}|=|z|$, so $|z|=0$ or $|z|=1$. The first gives $z=0$. In the second case $\bar{z}=1/z$, so the equation becomes $z^2=1/z$, that is $z^3=1$, contributing the three cube roots of unity. Four solutions in all.

The reason the count is not two is that $\bar{z}$ is not a polynomial function of $z$. Conjugation cannot be written using only additions and multiplications of $z$, so the fundamental theorem of algebra says nothing about an equation containing it.

The working rule for the whole chapter follows: an equation involving $\bar{z}$ or $|z|$ is a pair of real equations in disguise, and the first move is either to take moduli or to write $z=x+iy$ — never to count degrees.

1. The identity that does most of the work

Everything algebraic in this chapter comes from $|z{|}^2=z\bar{z}$. Expanding a modulus of a sum with it,

$|z_1+z_2{|}^2=(z_1+z_2)({\bar{z}}_1+{\bar{z}}_2)=|z_1{|}^2+|z_2{|}^2+2\mathrm{Re}(z_1{\bar{z}}_2).$

Replacing $z_2$ by $-z_2$ and adding gives the parallelogram law $|z_1+z_2{|}^2+|z_1-z_2{|}^2=2\left(|z_1{|}^2+|z_2{|}^2\right)$, which is worth recognising whenever a question supplies two of those three quantities.

The same expansion settles the equality case of the triangle inequality. Since $\mathrm{Re}(z_1{\bar{z}}_2)\le |z_1||z_2|$, we get $|z_1+z_2|\le |z_1|+|z_2|$, with equality exactly when $z_1{\bar{z}}_2$ is real and non-negative, that is when the two have the same argument or one of them is zero.

Illustration 1

Show that $|z_1+z_2|=|z_1-z_2|$ forces $z_1/z_2$ to be purely imaginary.

Squaring both sides and using the expansion, the two $|z_1{|}^2+|z_2{|}^2$ terms cancel and we are left with $4\mathrm{Re}(z_1{\bar{z}}_2)=0$. Dividing by $|z_2{|}^2$, this says $\mathrm{Re}(z_1/z_2)=0$. Geometrically the diagonals of the parallelogram are equal, so it is a rectangle and the sides are perpendicular.

Illustration 2

If $|z|=1$ and $z\ne \pm 1$, prove that $\frac{z-1}{z+1}$ is purely imaginary.

On the unit circle $\bar{z}=1/z$. Conjugating the expression,

$\overline{\left(\frac{z-1}{z+1}\right)}=\frac{\bar{z}-1}{\bar{z}+1}=\frac{\frac{1}{z}-1}{\frac{1}{z}+1}=\frac{1-z}{1+z}=-\frac{z-1}{z+1}.$

A number equal to the negative of its own conjugate has zero real part. The substitution $\bar{z}=1/z$ on the unit circle is the single most useful trick in the chapter and replaces several lines of $x$ and $y$ algebra.

2. Multiplication is rotation

Writing $z=r{e}^{i\theta }$, multiplying by $w=\rho e^{i\varphi }$ scales the length by $\rho$ and turns the picture by $\varphi$. Multiplication by $i$ is a quarter turn anticlockwise; multiplication by $e^{i\alpha }$ is a turn through $\alpha$.

Rotating a point $z$ through $\alpha$ about a centre $z_0$ therefore means translating the centre to the origin, turning, and translating back:

$z⟼z_0+(z-z_0)e^{i\alpha }.$

This is the tool that makes complex numbers a geometry method rather than an algebra topic, and it is the main reason Advanced sets configuration questions in the Argand plane at all.

One caution about arguments. The identity $\arg (z_1{z}_2)=\arg z_1+\arg z_2$ holds only up to a multiple of $2\pi$, because the principal argument is confined to $(-\pi ,\pi ]$. Adding two arguments near $\pi$ produces a value that must be reduced before it is reported.

Illustration 3

$A(1+2i)$ and $B(4+6i)$ are adjacent vertices of a square $ABCD$ lettered anticlockwise. Find $C$ and $D$.

The side vector is $B-A=3+4i$, of length $5$. Turning it a quarter turn anticlockwise multiplies by $i$, giving $i(3+4i)=-4+3i$. So

$C=B+(-4+3i)=i,D=A+(-4+3i)=-3+5i.$

As a check, $|D-C|=|-3+4i|=5$ and $D-C=-3+4i$ is $i$ times $B-A$ reversed, so the figure closes.

Illustration 4

Evaluate ${\left(\frac{1+i}{1-i}\right)}^{2026}$.

Rather than expanding, note $\frac{1+i}{1-i}=\frac{(1+i{)}^2}{2}=i$. So the answer is $i^{2026}=i^2=-1$, since $2026=4(506)+2$. Reducing a quotient to a single unit-modulus number before applying De Moivre's theorem is almost always faster than binomial expansion.

3. The roots of unity are a regular polygon

The solutions of $z^n=1$ are ${\omega }^k=e^{2\pi ik/n}$ for $k=0,1,\dots ,n-1$: $n$ equally spaced points on the unit circle. Two consequences carry most of the questions.

Summing the geometric series gives $1+\omega +{\omega }^2+\cdots +{\omega }^{n-1}=0$, and more generally ${\sum }_k{\omega }^{rk}=0$ unless $n$ divides $r$, in which case the sum is $n$. This is how a sum whose terms look unrelated collapses to zero.

Factorising, $z^n-1={\prod }_{k=0}^{n-1}(z-{\omega }^k)$. Dividing by $z-1$ and letting $z\to 1$,

${\prod }_{k=1}^{n-1}\left(1-{\omega }^k\right)=n.$

Illustration 5

A regular $n$-gon is inscribed in a unit circle. Show that the product of the distances from one vertex to all the others equals $n$.

Place the vertices at the $n$th roots of unity and take the vertex at $1$. The distances are $\left|1-{\omega }^k\right|$ for $k=1,\dots ,n-1$, and the product of those moduli is the modulus of the product, which is $n$ by the identity above. For a square this predicts $\sqrt{2}\cdot 2\cdot \sqrt{2}=4$, which is right.

Illustration 6

If $\omega \ne 1$ is a cube root of unity, evaluate $(1-\omega +{\omega }^2{)}^7+(1+\omega -{\omega }^2{)}^7$.

Since $1+\omega +{\omega }^2=0$, we have $1+{\omega }^2=-\omega$ and $1+\omega =-{\omega }^2$. So the brackets are $-2\omega$ and $-2{\omega }^2$, and the expression is

$-128\left({\omega }^7+{\omega }^{14}\right)=-128\left(\omega +{\omega }^2\right)=128.$

4. Roots of any complex number, and extremes of the modulus

De Moivre's theorem, ${\left(\cos \theta +i\sin \theta \right)}^n=\cos n\theta +i\sin n\theta$, is an identity for integer $n$ and a statement about one of several values for fractional $n$. That distinction is what makes the $n$th roots of a general complex number worth setting out separately.

To solve $z^n=w$ with $w=\rho e^{i\varphi }$, write the target in every equivalent form $\rho e^{i(\varphi +2\pi k)}$ and take the root of each:

$z_k={\rho }^{1/n}{e}^{i(\varphi +2\pi k)/n},k=0,1,\dots ,n-1.$

These are $n$ equally spaced points on a circle of radius ${\rho }^{1/n}$, so the roots of any complex number form a regular polygon, and for $n\ge 2$ they sum to zero for the same reason the roots of unity do. Only $k=0,\dots ,n-1$ are new; larger $k$ repeats the list.

A square root can also be extracted without any trigonometry, which is usually faster when the number is given in the form $a+ib$. Setting $\sqrt{a+ib}=x+iy$ and comparing parts gives $x^2-y^2=a$ and $2xy=b$, while taking moduli gives $x^2+y^2=\sqrt{a^2+b^2}$. Adding and subtracting the first and third of these isolates $x$ and $y$, and the sign of $b$ decides whether $x$ and $y$ share a sign.

Illustration 7

Find $\sqrt{3+4i}$.

Here $\left|3+4i\right|=5$, so $x^2=\frac{5+3}{2}=4$ and $y^2=\frac{5-3}{2}=1$. Since $b=4>0$, the product $xy$ is positive, so the two roots are $\pm (2+i)$. Squaring back, $(2+i{)}^2=4+4i-1=3+4i$, as required.

Illustration 8

If $\left|z-\frac{1}{z}\right|=2$, find the largest possible value of $|z|$.

The triangle inequality gives $|z|\le \left|z-\frac{1}{z}\right|+\frac{1}{|z|}$, so with $r=|z|$ we need $r\le 2+\frac{1}{r}$, that is $r^2-2r-1\le 0$ and hence $r\le 1+\sqrt{2}$.

This is only an upper bound until it is attained. Writing $z=r{e}^{i\theta }$ and expanding the modulus squared gives $r^2+\frac{1}{r^2}-2\cos 2\theta =4$, so the bound is reached when $\cos 2\theta =1$, that is when $z$ is real. Indeed $z=1+\sqrt{2}$ gives $z-\frac{1}{z}=\left(1+\sqrt{2}\right)-\left(\sqrt{2}-1\right)=2$. Producing the case of equality is part of the answer, not an optional check.

5. Loci, and why an argument condition gives an arc

Six standard conditions cover nearly every locus question, and the last is where marks are lost.

The last row is the trap. The set of points from which the segment $ab$ subtends a fixed angle $\alpha$ is a pair of arcs, one on each side of the line $ab$, and the two sides correspond to $\alpha$ and $\alpha -\pi$. Fixing the argument, rather than fixing only the size of the angle, selects one of the two arcs and excludes the endpoints $a$ and $b$ themselves.

Illustration 9

Identify the locus of $|z-1|=2|z-4|$.

Squaring with $z=x+iy$: $x^2-2x+1+y^2=4\left(x^2-8x+16+y^2\right)$, so $3x^2+3y^2-30x+63=0$, that is $x^2+y^2-10x+21=0$. This is a circle of centre $(5,0)$ and radius $\sqrt{25-21}=2$.

Note that $4$ lies inside it and $1$ lies outside. The Apollonius circle never passes through either of the two given points unless $k=1$, in which case it degenerates to the perpendicular bisector.

Illustration 10

Describe the locus $\arg \frac{z-1}{z+1}=\frac{\pi }{2}$.

The segment from $-1$ to $1$ subtends a right angle, so $z$ lies on the circle with that segment as diameter, namely $|z|=1$. But the argument is fixed at $+\pi /2$ rather than $\pm \pi /2$, which selects only the points above the real axis. The locus is the open upper semicircle, endpoints excluded. Reporting the full circle is the standard error.

6. Triangles in the Argand plane

The quotient $\frac{z_3-z_2}{z_1-z_2}$ has modulus equal to the ratio of the two side lengths at $z_2$ and argument equal to the angle at $z_2$ measured from $z_2{z}_1$ to $z_2{z}_3$. Every triangle condition follows from reading that one quotient.

If the quotient is real the three points are collinear; if it is purely imaginary the angle at $z_2$ is a right angle; if it equals $e^{\pm i\pi /3}$ with modulus $1$ the triangle is equilateral.

Illustration 11

Derive the equilateral condition $z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1$.

The condition is equivalent to $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0$: clearing denominators in the second form produces exactly $\left(z_1{z}_2+z_2{z}_3+z_3{z}_1\right)-\left(z_1^2+z_2^2+z_3^2\right)$. Now if the triangle is equilateral, rotating $z_3-z_1$ about $z_1$ by $\pm 60^{\circ }$ carries it to $z_2-z_1$, and the resulting relation between the three differences gives precisely that vanishing sum. The symmetric form is easier to test, but the rotation form is what you use to construct the third vertex.

Illustration 12

If $z_1=1$, $z_2=4+3i$ and the triangle $z_1{z}_2{z}_3$ is equilateral with $z_3$ on the left of the directed side from $z_1$ to $z_2$, find $z_3$.

Rotate $z_2$ about $z_1$ through $+60^{\circ }$:

$z_3=1+(3+3i)\left(\cos 60^{\circ }+i\sin 60^{\circ }\right)=1+(3+3i)\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right).$

Expanding, $(3+3i)\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=\frac{3-3\sqrt{3}}{2}+\frac{3+3\sqrt{3}}{2}i$, so $z_3=\frac{5-3\sqrt{3}}{2}+\frac{3+3\sqrt{3}}{2}i$. The modulus of $z_3-z_1$ is unchanged at $|3+3i|=3\sqrt{2}$, as a rotation demands.

7. Quadratics whose coefficients are not real

The habit that non-real roots come in conjugate pairs is a theorem about real coefficients, and it is proved by conjugating the equation, which only reproduces the original when every coefficient equals its own conjugate. With complex coefficients the roots are unrelated.

Illustration 13

Solve $z^2-(3+i)z+(2+2i)=0$.

Trying $z=2$ gives $4-6-2i+2+2i=0$, so $2$ is a root, and the sum of the roots is $3+i$, making the other root $1+i$. The two are not conjugates, and neither is real, even though one of them happens to be. Nothing is wrong: the conjugate-pair theorem simply does not apply.

Two companion theorems have the same shape and the same fine print. Non-real roots pair as conjugates when every coefficient is real, and irrational surd roots pair as $p\pm \sqrt{q}$ when every coefficient is rational. Both are proved by applying a map that fixes the coefficients — conjugation in the first case, the substitution $\sqrt{q}\mapsto -\sqrt{q}$ in the second — and both collapse if a single coefficient falls outside the required field.

Running the relations backwards forms a quadratic from its roots: given $\alpha$ and $\beta$, the monic equation is $x^2-\left(\alpha +\beta \right)x+\alpha \beta =0$. This is how questions are set that supply one root and require the other, and it is why a single non-real root together with the phrase "real coefficients" is enough information to reconstruct the whole equation.

For real coefficients the standard relations still govern everything. For $a{x}^2+bx+c=0$ the sum is $-b/a$ and the product $c/a$, and the two most useful derived quantities are ${\alpha }^2+{\beta }^2={\left(\alpha +\beta \right)}^2-2\alpha \beta$ and $(\alpha -\beta {)}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta$.

Illustration 14

Find the condition for $x^2+bx+c=0$ and $x^2+b^{\prime }x+c^{\prime }=0$ to have a common root.

Subtracting the two equations kills the $x^2$ term and leaves $(b-b^{\prime })x+(c-c^{\prime })=0$, so the common root must be $x=\frac{c^{\prime }-c}{b-b^{\prime }}$ when $b\ne b^{\prime }$. Substituting into either equation gives the condition

${\left(c^{\prime }-c\right)}^2+b\left(c^{\prime }-c\right)\left(b-b^{\prime }\right)+c{\left(b-b^{\prime }\right)}^2=0.$

Subtracting first is always the right opening move, because it reduces the problem from two quadratics to one linear equation.

8. Where the roots of a real quadratic lie

Advanced asks for the values of a parameter that place both roots in an interval far more often than it asks for the roots themselves. With $f(x)=a{x}^2+bx+c$ and real coefficients, every such question is answered by three ingredients: the discriminant, the sign of $af(k)$ at each boundary, and the position of the vertex $-b/2a$.

The factor $a$ appears because the parabola opens downwards when $a<0$, which reverses every sign statement about $f$.

One further family uses the same picture without mentioning roots at all. A real quadratic $a{x}^2+bx+c$ keeps one sign for every real $x$ exactly when it has no real root, so $a{x}^2+bx+c>0$ for all $x$ requires $a>0$ together with $D<0$, and the reversed inequality requires $a<0$ with $D<0$.

When the parameter sits in the leading coefficient, the case $a=0$ falls outside this argument entirely and has to be tested on its own, exactly as the vanishing leading coefficient did in the range problems of the previous chapter.

A question that asks for all $k$ making $\left(k-1\right)x^2+2kx+k+3$ positive for every $x$ has $k=1$ as a genuine candidate, since the expression then becomes the linear $2x+4$, which is not always positive and so is rejected — but rejected for a reason, not by omission.

Illustration 15

For which real $\lambda$ do both roots of $x^2-2\lambda x+{\lambda }^2-1=0$ lie in $(-2,4)$?

The quadratic is $(x-\lambda {)}^2-1$, so its roots are $\lambda \pm 1$ and the discriminant is always positive. Requiring $\lambda -1>-2$ and $\lambda +1<4$ gives $-1<\lambda <3$.

Checking against the general conditions: $f(-2)={\lambda }^2+4\lambda +3=(\lambda +1)(\lambda +3)>0$ and $f(4)={\lambda }^2-8\lambda +15=(\lambda -3)(\lambda -5)>0$ and $-2<\lambda <4$. Intersecting these gives $-1<\lambda <3$ as well, which is the check worth doing whenever the roots are not this easy to see.

Summary

An equation containing $\bar{z}$ or $|z|$ is not a polynomial equation, so degree counting does not apply: $z^2=\bar{z}$ has four solutions. Take moduli first, or split into real and imaginary parts. The identity $|z{|}^2=z\bar{z}$ generates the expansion $|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2+2\mathrm{Re}(z_1{\bar{z}}_2)$, and with it the parallelogram law and the equality case of the triangle inequality. On the unit circle, $\bar{z}=1/z$ replaces most coordinate algebra.

Multiplication by $e^{i\alpha }$ rotates, so a rotation about $z_0$ is $z\mapsto z_0+(z-z_0)e^{i\alpha }$, and that single formula constructs squares and equilateral triangles directly. The $n$th roots of unity are a regular polygon: they sum to zero, and the product of the distances from one vertex to the rest is $n$.

Among loci, only the argument condition needs care, because it gives one arc rather than a full circle and excludes the two base points. The Apollonius circle passes through neither given point, and degenerates to the perpendicular bisector when the ratio is one.

For quadratics, the conjugate-pair theorem needs real coefficients and fails without them. Common-root questions start by subtracting the equations. Root-location questions are answered by the discriminant, the sign of $af(k)$ at each boundary, and the position of the vertex — never by solving for the roots when the parameter is still present.