Complex Numbers

1.Cartesian form, the conjugate, and division

A complex number is written $z=a+bi$ with $a,b$ real and $i^{2}=-1$. The real number $a=\operatorname{Re} z$ is the real part and $b=\operatorname{Im} z$ is the imaginary part; note that $\operatorname{Im} z=b$ is a real number, not $bi$. Each $z$ is a point $(a,b)$ of the plane, and addition is componentwise, so the geometry of addition is ordinary vector addition. Multiplication is where the new structure lives, and it is the reason complex numbers are useful.

Multiplication follows from the distributive law and $i^{2}=-1$: $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. This is the FOIL expansion with one extra rule. Students should practice expanding products until the pattern is automatic, since almost every computation in this topic begins with a product of this kind, and a sign error from $i^{2}$ is the most common source of wrong answers.

The conjugate of $z=a+bi$ is $\bar z=a-bi$. Conjugation is the reflection of the plane across the real axis, and it respects arithmetic: $\overline{z+w}=\bar z+\bar w$ and $\overline{zw}=\bar z\,\bar w$. Two identities follow at once: $\operatorname{Re} z=\frac{z+\bar z}{2}$ and $\operatorname{Im} z=\frac{z-\bar z}{2i}$. These let one extract real and imaginary parts from a formula without expanding it into coordinates.

The key identity is $z\bar z=a^{2}+b^{2}$. Proof: $(a+bi)(a-bi)=a^{2}-(bi)^{2}=a^{2}+b^{2}$. The right side is the square of the distance from $z$ to the origin, which we write $|z|^{2}$. So $|z|^{2}=z\bar z$, and this single formula converts modulus questions into algebra. It also shows that $|zw|^{2}=zw\,\bar z\bar w=|z|^{2}|w|^{2}$, hence $|zw|=|z|\,|w|$ for all $z,w$.

Division is handled by multiplying numerator and denominator by the conjugate of the denominator. If $w\ne0$ then $\frac{z}{w}=\frac{z\bar w}{w\bar w}=\frac{z\bar w}{|w|^{2}}$, and the new denominator is a positive real number. For instance $\frac{3+4i}{1-2i}=\frac{(3+4i)(1+2i)}{5}=\frac{-5+10i}{5}=-1+2i$. The method works for every nonzero $w$, and it is the standard way to simplify any quotient whose denominator is not real.

Why does multiplying by $\bar w/\bar w$ not change the value? Because the fraction equals one, and the denominator $w\bar w$ is real. Multiplying by $i$ or by a real number instead would leave the denominator non-real, so the conjugate is the correct multiplier. Once the denominator is real, the real and imaginary parts of the quotient can be read off directly, with no further work.

Two beginner traps dominate this lesson. First, students carry the wrong sign through a product because they forget that $i^{2}=-1$. Second, students try to clear a complex denominator by multiplying it by $i$, which does not make it real. A useful habit is to write the final answer in the form $x+yi$ and check each part separately, because a correct modulus can hide a wrong real part.

• $i^{2}=-1$ and $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$
• $\overline{a+bi}=a-bi$, $\overline{zw}=\bar z\,\bar w$
• $z\bar z=|z|^{2}=a^{2}+b^{2}$
• $\dfrac{z}{w}=\dfrac{z\bar w}{|w|^{2}}$ for $w\ne0$
• $\operatorname{Re} z=\dfrac{z+\bar z}{2}$, $\operatorname{Im} z=\dfrac{z-\bar z}{2i}$

• Treating $\operatorname{Im} z$ as $bi$ rather than the real number $b$.
• Dividing by a complex number without multiplying by its conjugate, which leaves $i$ in the denominator.
• Expanding $(a+bi)^{2}$ and dropping the cross term $2abi$.

2.Modulus, argument, and polar form

Every nonzero $z=a+bi$ has polar coordinates: the modulus $r=|z|=\sqrt{a^{2}+b^{2}}$ and an argument $\theta=\arg z$, the angle measured counterclockwise from the positive real axis. Then $a=r\cos\theta$ and $b=r\sin\theta$, so $z=r(\cos\theta+i\sin\theta)$. The argument is defined only modulo $2\pi$; the principal value is usually taken in $(-\pi,\pi]$, and the problem should say which convention it uses.

Multiplication in polar form is where geometry becomes algebra. Let $z=r(\cos\alpha+i\sin\alpha)$ and $w=s(\cos\beta+i\sin\beta)$. Multiplying out and applying the addition formulas, the real part of $zw/(rs)$ is $\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta)$ and the imaginary part is $\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta)$. Hence $zw=rs\big(\cos(\alpha+\beta)+i\sin(\alpha+\beta)\big)$: moduli multiply and arguments add.

Euler's notation $e^{i\theta}=\cos\theta+i\sin\theta$, justified in a later lesson, packages this as $re^{i\alpha}\cdot se^{i\beta}=rse^{i(\alpha+\beta)}$, which is just the law of exponents. Division gives $\frac{z}{w}=\frac{r}{s}e^{i(\alpha-\beta)}$. With practice you should read the modulus and argument of any product or quotient immediately, without expanding anything.

Converting back to Cartesian form uses a reference angle. For $z=-1+i\sqrt3$ we have $r=2$, and $\cos\theta=-\tfrac12$, $\sin\theta=\tfrac{\sqrt3}{2}$, so $\theta=\tfrac{2\pi}{3}$. The reference angle is found from $\arctan\left|\frac{b}{a}\right|$, and then the quadrant is fixed by the signs of $a$ and $b$. Quadrant errors are common because $\arctan$ alone cannot distinguish $\theta$ from $\theta+\pi$.

The modulus is the distance from $z$ to $0$, so $|z-w|$ is the distance between two points. Many contest problems start by rewriting a condition such as $|z|=1$ or $\arg z=\pi/3$, after which the problem is about a circle or a ray from the origin. Learning to recognize $|z-c|=r$ as a circle and $\arg(z-c)=\theta$ as a ray from $c$ is one of the most useful habits in this topic.

Modulus is multiplicative and argument is additive, so powers are easy. For example $|(1+i)^{10}|=(\sqrt2)^{10}=32$, and $\arg(1+i)^{10}=10\cdot\frac{\pi}{4}=\frac{5\pi}{2}\equiv\frac{\pi}{2}$, so $(1+i)^{10}=32i$. Check directly: $(1+i)^{2}=2i$, and $(2i)^{5}=32i^{5}=32i$.

Two mistakes are common. First, students use $\arg(a+bi)=\arctan(b/a)$ without checking the quadrant. Second, students forget that $\arg$ is only defined up to multiples of $2\pi$, so $-30^\circ$ and $330^\circ$ name the same direction. Always state which representative of the argument you are using before comparing two answers.

• $z=r(\cos\theta+i\sin\theta)$ with $r=|z|$ and $\theta=\arg z$
• $|zw|=|z|\,|w|$ and $\arg(zw)\equiv\arg z+\arg w \pmod{2\pi}$
• $\left|\dfrac{z}{w}\right|=\dfrac{|z|}{|w|}$ and $|z|^{2}=z\bar z$
• $\arg\bar z\equiv-\arg z \pmod{2\pi}$

• Reading the argument off $\arctan(b/a)$ without checking which quadrant the point lies in.
• Forgetting that the argument is defined only modulo $2\pi$, so two different-looking answers may be equal.
• Using $r\,e^{i\theta}$ with $\theta$ in degrees in one place and radians in another.

3.De Moivre's theorem and integer powers

De Moivre's theorem states $(\cos\theta+i\sin\theta)^{n}=\cos n\theta+i\sin n\theta$ for every integer $n$. For $n\ge0$ it follows by induction. The case $n=1$ is trivial. If the statement holds for $n$, then $(\cos\theta+i\sin\theta)^{n+1}=(\cos n\theta+i\sin n\theta)(\cos\theta+i\sin\theta)$, and the product rule from the previous lesson gives $\cos(n+1)\theta+i\sin(n+1)\theta$.

Negative exponents follow from the observation that $(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)=\cos^{2}\theta+\sin^{2}\theta=1$, so $(\cos\theta+i\sin\theta)^{-1}=\cos\theta-i\sin\theta=\cos(-\theta)+i\sin(-\theta)$. The formula therefore holds for all integers $n$. Geometrically, multiplying by a unit complex number rotates the plane, and performing that rotation $n$ times is a single rotation by $n\theta$.

For a general nonzero complex number write $z=r(\cos\theta+i\sin\theta)$. Then $z^{n}=r^{n}(\cos n\theta+i\sin n\theta)$. Two things happen at once: the modulus is raised to the power $n$, and the argument is multiplied by $n$ (modulo $2\pi$). This is the single most useful consequence in competition work, because an integer power of a complicated expression often reduces to a short computation with an angle.

Example. Let $z=\sqrt3+i$. Then $r=2$ and $\theta=\frac{\pi}{6}$, so $z^{6}=2^{6}\big(\cos\pi+i\sin\pi\big)=-64$. Nothing was expanded. Repeated squaring and multiplying would also work, but it takes many steps and invites arithmetic slips, while the polar computation takes one line once the modulus and argument are known.

De Moivre also produces trigonometric identities. Expanding $(\cos\theta+i\sin\theta)^{3}$ by the binomial theorem gives $\cos^{3}\theta-3\cos\theta\sin^{2}\theta+i\,(3\cos^{2}\theta\sin\theta-\sin^{3}\theta)$. Comparing real parts and using $\sin^{2}\theta=1-\cos^{2}\theta$ gives $\cos3\theta=4\cos^{3}\theta-3\cos\theta$, and comparing imaginary parts gives $\sin3\theta=3\sin\theta-4\sin^{3}\theta$. Students should be able to reproduce both identities from the binomial expansion.

The same idea solves equations. The solutions of $z^{n}=w$, where $w=\rho(\cos\varphi+i\sin\varphi)$, are $z_{k}=\rho^{1/n}\big(\cos\frac{\varphi+2\pi k}{n}+i\sin\frac{\varphi+2\pi k}{n}\big)$ for $k=0,1,\dots,n-1$. Each $z_{k}$ satisfies the equation by De Moivre's theorem, and there are exactly $n$ distinct values, which matches the degree of $z^{n}-w$.

Common mistakes are applying De Moivre to $\cos\theta+i\sin\theta$ when the modulus is not $1$, so that the factor $r^{n}$ is lost, and reducing the angle carelessly when $n\theta$ is large. Reduce the angle modulo $2\pi$ before evaluating any cosine or sine, and keep track of the modulus separately.

• $(\cos\theta+i\sin\theta)^{n}=\cos n\theta+i\sin n\theta$ for all integers $n$
• $\big(r(\cos\theta+i\sin\theta)\big)^{n}=r^{n}(\cos n\theta+i\sin n\theta)$
• $\cos3\theta=4\cos^{3}\theta-3\cos\theta$
• $\sin3\theta=3\sin\theta-4\sin^{3}\theta$

• Applying De Moivre to a number whose modulus is not $1$ and forgetting the factor $r^{n}$.
• Expanding a high power by hand when the polar form would finish in one step.
• Using an angle $n\theta$ without reducing modulo $2\pi$ before reading off the cosine.

4.Euler's formula and exponential sums

Define $e^{i\theta}=\cos\theta+i\sin\theta$. The reason for this definition is that the exponential series $e^{x}=\sum_{n\ge0}\frac{x^{n}}{n!}$ converges for complex $x$, and $e^{x+y}=e^{x}e^{y}$ continues to hold. Substituting $x=i\theta$ and using $i^{2}=-1$, the even powers give $\sum_{m}\frac{(-1)^{m}\theta^{2m}}{(2m)!}=\cos\theta$ and the odd powers give $i\sum_{m}\frac{(-1)^{m}\theta^{2m+1}}{(2m+1)!}=i\sin\theta$. This is Euler's formula.

The consequences are immediate. The identity $e^{i\pi}=-1$ is Euler's identity. The rule $e^{i\theta}e^{i\varphi}=e^{i(\theta+\varphi)}$ is the angle-addition formula, and De Moivre's theorem is $(e^{i\theta})^{n}=e^{in\theta}$. Since $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$, every trigonometric polynomial becomes a polynomial in $e^{i\theta}$ and $e^{-i\theta}$, and algebra takes over.

The most important contest use is summing trigonometric series. Let $\omega=e^{i\theta}$ with $\omega\ne1$. The finite geometric series gives $\sum_{k=0}^{n-1}\omega^{k}=\frac{\omega^{n}-1}{\omega-1}$. If $\omega^{n}=1$, the numerator vanishes while the denominator does not, so the whole sum is $0$. In general the real part of this quotient gives a closed form for a cosine sum with arithmetic arguments.

To get the closed form, note that $\omega^{n}-1=e^{in\theta/2}\cdot2i\sin\frac{n\theta}{2}$ and $\omega-1=e^{i\theta/2}\cdot2i\sin\frac{\theta}{2}$. Therefore $\frac{\omega^{n}-1}{\omega-1}=e^{i(n-1)\theta/2}\,\frac{\sin(n\theta/2)}{\sin(\theta/2)}$. Taking real parts gives $\sum_{k=0}^{n-1}\cos k\theta=\cos\frac{(n-1)\theta}{2}\cdot\frac{\sin(n\theta/2)}{\sin(\theta/2)}$ whenever $\sin(\theta/2)\ne0$. The complex method is far cleaner than the product-to-sum manipulations it replaces.

Example: $\sum_{k=0}^{9}\cos\frac{k\pi}{5}$. Here $\omega=e^{i\pi/5}$ satisfies $\omega^{10}=1$ and $\omega\ne1$, so $\sum_{k=0}^{9}\omega^{k}=0$. Its real part is exactly the cosine sum, which therefore vanishes. The closed form agrees: $\sin\frac{10\pi}{10}=\sin\pi=0$.

Euler's formula also converts products of trigonometric functions into sums. Expanding $\cos^{n}\theta=\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^{n}$ with the binomial theorem and regrouping exponents gives a sum of cosines of multiples of $\theta$. This is the same technique used in the trigonometric identities lesson, and it avoids the long chains of double-angle formulas that students otherwise write down.

Pitfalls: confusing $e^{i\theta}$ with a real number that grows with $\theta$ (it is a point on the unit circle); dividing by $\omega-1$ when $\omega=1$; and using the closed form when $\sin(\theta/2)=0$, in which case every term of the sum is $1$ or $\cos$ of a multiple of $2\pi$ and the sum is simply $n$ or the count of such terms.

• $e^{i\theta}=\cos\theta+i\sin\theta$ and $e^{i\pi}=-1$
• $\cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}{2}$ and $\sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$
• $\sum_{k=0}^{n-1}\omega^{k}=\dfrac{\omega^{n}-1}{\omega-1}$ for $\omega\ne1$
• $\sum_{k=0}^{n-1}\cos k\theta=\cos\dfrac{(n-1)\theta}{2}\cdot\dfrac{\sin(n\theta/2)}{\sin(\theta/2)}$

• Dividing by $\omega-1$ without first checking that $\omega\ne1$.
• Taking the real part of $\sum\omega^{k}$ but mis-indexing the sum, so that the number of terms is off by one.
• Applying the closed form in the degenerate case $\sin(\theta/2)=0$.

5.Roots of unity: sums, products, and factorization

An $n$th root of unity is a solution of $z^{n}=1$. Writing $z=re^{i\theta}$ forces $r=1$ and $n\theta\in2\pi\mathbb Z$, so the solutions are $\omega_k=e^{2\pi ik/n}$ for $k=0,1,\dots,n-1$. There are exactly $n$ of them, equally spaced on the unit circle, and they form a group under multiplication. A root is primitive when its order is exactly $n$, which happens for $\omega_k$ with $\gcd(k,n)=1$.

The factorization $x^{n}-1=\prod_{k=0}^{n-1}(x-\omega_k)$ holds identically in $x$, because both sides are monic of degree $n$ and have the same $n$ distinct roots. Substituting $x=2$ gives $\prod_{k=0}^{n-1}(2-\omega_k)=2^{n}-1$, a product of complex numbers whose value is an integer. This is the typical way a product over roots of unity is evaluated: identify a polynomial with those roots, and evaluate it.

Vieta's formulas give the symmetric functions. The sum of all $n$th roots is $0$ for $n\ge2$, because the coefficient of $x^{n-1}$ in $x^{n}-1$ is zero. The product of all $n$th roots is $(-1)^{n+1}$, because for $x^{n}-1$ the product of the roots equals $(-1)^{n}$ times the constant term $-1$. Roots also come in conjugate pairs $\omega_k,\omega_{-k}$, which is why the real parts of many sums are symmetric.

The orthogonality rule is the most used fact in this topic. For an integer $m$, $\sum_{k=0}^{n-1}\omega_k^{m}$ equals $n$ if $n\mid m$ and $0$ otherwise. Indeed $\omega_k^{m}=\omega_1^{mk}$ is a root of unity, and a geometric series of roots of unity vanishes unless its ratio is $1$. Many problems reduce to sums of this form after exponents are reduced modulo $n$.

Logarithmic differentiation gives a reciprocal sum. Differentiating $x^{n}-1=\prod_{k}(x-\omega_k)$ and dividing by the original gives $\sum_{k=0}^{n-1}\frac{1}{x-\omega_k}=\frac{nx^{n-1}}{x^{n}-1}$. At $x=2$ with $n=7$ this yields $\sum_{k=0}^{6}\frac{1}{2-\omega_k}=\frac{7\cdot2^{6}}{2^{7}-1}=\frac{448}{127}$. The identity is valid at every $x$ that is not a root, so it is more flexible than most sum formulas.

Moduli of differences are also tractable. Since $\bar\omega=\omega^{-1}$, $|1-\omega|^{2}=(1-\omega)(1-\omega^{-1})=2-\omega-\omega^{-1}=2-2\cos\theta$. Summing over all $n$th roots of unity gives $\sum_{k=0}^{n-1}|1-\omega_k|^{2}=2n$. Also $\prod_{k=1}^{n-1}(1-\omega_k)=\lim_{x\to1}\frac{x^{n}-1}{x-1}=n$, so the product of the distances from $1$ to the other roots is $n$.

Pitfalls: forgetting that $\omega_0=1$ is itself a root, so sums over all $n$th roots include it, and dividing by $1-\omega_k$ when $k=0$, which is zero. When a problem uses a root $\omega$ of a particular order, first write each exponent modulo that order; the reduced problem is usually much smaller.

• $\omega_k=e^{2\pi ik/n}$ for $k=0,\dots,n-1$ are the $n$th roots of unity
• $x^{n}-1=\prod_{k=0}^{n-1}(x-\omega_k)$, so $\prod_{k=0}^{n-1}(x-\omega_k)$ at $x=2$ is $2^{n}-1$
• $\sum_{k=0}^{n-1}\omega_k^{m}=n$ if $n\mid m$, and $0$ otherwise
• $\sum_{k=0}^{n-1}\dfrac{1}{x-\omega_k}=\dfrac{nx^{n-1}}{x^{n}-1}$
• $|1-\omega|^{2}=(1-\omega)(1-\omega^{-1})=2-2\cos\theta$

• Excluding the root $1$ (that is, $\omega_0$) from a sum over all roots.
• Evaluating $\omega^{m}$ without first reducing $m$ modulo $n$.
• Using the sum $\sum_k\omega_k^{m}$ formula when $n$ does not divide $m$ and then forgetting the sum is zero.

6.Rotations and similarity in the complex plane

Multiplication by a unit complex number $u=e^{i\varphi}$ is a rotation of the plane about the origin through angle $\varphi$, because $|uz|=|z|$ and $\arg(uz)=\arg z+\varphi$. Multiplication by a positive real $t$ is a dilation. Multiplication by $re^{i\varphi}$ is a spiral similarity: rotate by $\varphi$ and scale by $r$. Every orientation-preserving similarity of the plane has this form, up to a translation.

To rotate $z$ by $\varphi$ about a general center $c$, translate by $-c$, rotate, and translate back: $w=c+e^{i\varphi}(z-c)$. The vector $z-c$ points from the center to $z$; rotating that vector gives $e^{i\varphi}(z-c)$, and adding $c$ moves it back to the correct place. This is the most efficient way to handle any rotation about a point, and it avoids coordinate computations with sines and cosines.

Example. Rotate $z=4$ by $60^\circ$ counterclockwise about $c=2$. Then $w=2+e^{i\pi/3}(4-2)=2+2\left(\tfrac12+\tfrac{\sqrt3}{2}i\right)=3+i\sqrt3$. A quick check is that $|w-2|=|z-2|=2$, which is the statement that a rotation preserves distance from its center.

Equilateral triangles have a clean algebraic test. If $a,b,c$ are the vertices of a triangle, then $c-a=e^{\pm i\pi/3}(b-a)$ exactly when the triangle is equilateral, with the sign fixed by orientation. Since $\zeta=e^{i\pi/3}$ satisfies $\zeta^{2}-\zeta+1=0$, one can show that $a,b,c$ form an equilateral triangle if and only if $a^{2}+b^{2}+c^{2}=ab+bc+ca$. This condition is symmetric in $a,b,c$ and contains no square roots, so it is usually preferable to a distance computation.

Similarity and angles are captured by ratios. The quantity $\frac{c-a}{b-a}$ does not change under a translation, rotation, or dilation applied to the whole triangle, so two triangles are directly similar exactly when these ratios agree. For an equilateral triangle the apex ratio is $e^{\pm i\pi/3}$; for an isosceles right triangle with the right angle at $a$ it is $\pm i$.

Perpendicularity is an imaginary ratio. For nonzero vectors $u,v$, the quotient $\frac{u}{v}=\frac{u\bar v}{|v|^{2}}$ has real part equal to $\frac{\operatorname{Re}(u\bar v)}{|v|^{2}}$, and $\operatorname{Re}(u\bar v)$ is the dot product of $u$ and $v$. Therefore $u\perp v$ if and only if $\frac{u}{v}$ is purely imaginary. Multiplication by $i$ is rotation by $90^\circ$, which is the same fact seen geometrically.

Pitfalls: writing a rotation about $c$ as $z e^{i\varphi}$, which rotates about the origin instead; and confusing counterclockwise with clockwise, since the sign of the exponent decides the orientation. Test any formula on a single easy point, such as $z=c+1$, before trusting it.

• $z\mapsto uz$ with $|u|=1$ is a rotation about $0$ through $\arg u$
• $w=c+e^{i\varphi}(z-c)$ rotates $z$ about $c$ through $\varphi$
• $u\perp v \iff u/v\in i\mathbb R$
• $a,b,c$ are equilateral $\iff a^{2}+b^{2}+c^{2}=ab+bc+ca$

• Rotating about the origin when the problem specifies a different center.
• Confusing the counterclockwise and clockwise rotations, which differ by the sign of the angle.
• Using a distance formula for an equilateral condition when the symmetric identity is shorter.

7.Loci, distances, and the triangle inequality

Distances are moduli: $|z-w|$ is the distance between the points $z$ and $w$. A condition on a modulus defines a familiar curve. The set $|z-c|=r$ is a circle with center $c$ and radius $r$; the set $|z-a|=|z-b|$ is the perpendicular bisector of the segment $ab$; and $\arg(z-c)=\theta$ is a ray from $c$. Converting a geometric condition into a modulus equation is the first step in nearly every loci problem.

The perpendicular bisector has a short derivation. Using $|z-a|^{2}=|z|^{2}-2\operatorname{Re}(z\bar a)+|a|^{2}$, the equation $|z-a|=|z-b|$ becomes $2\operatorname{Re}\big(z(\bar b-\bar a)\big)=|b|^{2}-|a|^{2}$. This is a linear equation in the coordinates of $z$, so the locus is a line. Its normal direction is $b-a$, so it is perpendicular to $ab$, and $z=\frac{a+b}{2}$ satisfies it, so it passes through the midpoint.

Extremal distances on a circle follow from the triangle inequality. If $|z|=r$ and $c$ is fixed, then $\big||c|-r\big|\le|z-c|\le r+|c|$. Both bounds are attained: the upper bound when $z$ and $-c$ point in the same direction, and the lower bound when $z$ and $c$ point in the same direction. So the maximum distance from a circle to a point is the radius plus the distance from the center to the point, and the minimum is the absolute difference.

The triangle inequality $|z+w|\le|z|+|w|$ has a precise equality case. Expanding gives $|z+w|^{2}=|z|^{2}+|w|^{2}+2\operatorname{Re}(z\bar w)$, and $\operatorname{Re}(z\bar w)\le|z\bar w|=|z|\,|w|$. Equality holds exactly when $z\bar w$ is a nonnegative real, which means $z$ and $w$ have the same argument. This characterization converts an inequality into a condition on angles, which is how maxima are usually located.

Apollonius circles are the loci $|z-a|=k|z-b|$ with $k\ne1$. Expanding in coordinates shows that the locus is a circle, whose center lies on the line through $a$ and $b$. A fast way to find the extreme values of $|z|$ on such a circle is the ray method: write $z=te^{i\varphi}$, substitute into the defining equation, and solve the resulting quadratic in $t$ for each direction $\varphi$.

A concrete check: the points $z$ with $|z-4|=2$ form a circle of radius $2$ centered at $4$. The closest point to the origin is $2$ units from $0$, because the center is $4$ units away and $4-2=2$. This is the same computation as the extremal-distance rule.

Pitfalls: confusing a locus $|z-a|=|z-b|$, which is a line, with $|z-a|+|z-b|=\text{const}$, which is an ellipse; and maximizing $|z-c|$ on a circle without checking that the optimal point is on that circle. Compare the candidate values carefully, including the endpoints of any segment that appears.

• $|z-c|=r$ is a circle; $|z-a|=|z-b|$ is the perpendicular bisector of $ab$
• $|z+w|\le|z|+|w|$, with equality iff $z$ and $w$ have the same argument
• For $|z|=r$: $\max|z-c|=r+|c|$ and $\min|z-c|=\big||c|-r\big|$
• $|z-a|^{2}=|z|^{2}-2\operatorname{Re}(z\bar a)+|a|^{2}$

• Calling the locus $|z-a|=|z-b|$ a circle rather than the perpendicular bisector.
• Using $|z|+|c|$ as the maximum of $|z-c|$ without requiring $z$ to lie on the circle in the correct direction.
• Forgetting that the locus $|z-a|+|z-b|=k$ is an ellipse, not a line segment, unless $k=|a-b|$.

8.Polynomials with complex roots

The fundamental theorem of algebra says that every nonconstant polynomial with complex coefficients has a complex root. By induction, a monic polynomial of degree $n$ factors as $P(x)=\prod_{k=1}^{n}(x-r_k)$, with exactly $n$ roots counted with multiplicity. This is why complex numbers are the natural setting for polynomial problems: no roots are missing from the count. Over the reals the situation is different, because irreducible quadratic factors can appear.

Vieta's formulas express the coefficients through the roots. Expanding $\prod_{k}(x-r_k)$ and comparing coefficients for a monic polynomial $x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}$ gives $\sum_k r_k=-a_{n-1}$, $\sum_{j<k}r_jr_k=a_{n-2}$, and $\prod_k r_k=(-1)^{n}a_0$. These identities are the main computational tool: they compute symmetric functions of the roots without ever finding the roots.

If $P$ has real coefficients and $P(r)=0$, then $P(\bar r)=\overline{P(r)}=0$, because conjugation respects sums and products. Hence the nonreal roots of a real polynomial come in conjugate pairs $r,\bar r$, and $(x-r)(x-\bar r)=x^{2}-2\operatorname{Re}(r)\,x+|r|^{2}$ has real coefficients. A real cubic therefore has either three real roots or one real root together with a conjugate pair.

Example. The polynomial $P(x)=x^{3}-3x^{2}+4x-2$ has the root $1+i$, since $(1+i)^{3}-3(1+i)^{2}+4(1+i)-2=(-2+2i)-6i+(4+4i)-2=0$. Its conjugate $1-i$ is also a root. Vieta gives $(1+i)+(1-i)+r=3$, so the third root is $r=1$. A check: $(1+i)(1-i)\cdot1=2=-(-2)$, matching the constant term.

Roots of unity factor $x^{n}\pm1$ over $\mathbb C$. The roots of $x^{n}+1$ are $e^{i\pi(2k+1)/n}$ for $k=0,\dots,n-1$. Over the reals, conjugate roots pair into real quadratics. For instance $x^{4}+1=(x^{2}+\sqrt2\,x+1)(x^{2}-\sqrt2\,x+1)$, because the roots $e^{\pm i\pi/4}$ and $e^{\pm3i\pi/4}$ pair up with sums $\pm\sqrt2$ and products $1$.

Power sums are computed from the coefficients. The sum of squares of the roots is $p_{2}=\left(\sum r_k\right)^{2}-2\sum_{j<k}r_jr_k$, which is $a_{n-1}^{2}-2a_{n-2}$ for a monic polynomial. Higher power sums follow by Newton's identities. This is the right tool when a problem asks for $\sum r_k^{m}$ and the roots themselves are inconvenient.

Pitfalls: assuming a real cubic with a nonreal root must have three real roots (it has exactly one real root, with the nonreal pair as its conjugate partners); forgetting the conjugate root when applying Vieta; and computing a square root of a negative discriminant without writing $\sqrt{-d}=i\sqrt d$ for $d>0$.

• $P(x)=\prod_{k=1}^{n}(x-r_k)$ for monic $P$ of degree $n$ (fundamental theorem of algebra)
• $\sum r_k=-a_{n-1}$, $\sum_{j<k}r_jr_k=a_{n-2}$, $\prod r_k=(-1)^{n}a_0$
• Real coefficients: $P(r)=0 \Rightarrow P(\bar r)=0$
• $x^{4}+1=(x^{2}+\sqrt2\,x+1)(x^{2}-\sqrt2\,x+1)$
• $\sum r_k^{2}=\left(\sum r_k\right)^{2}-2\sum_{j<k}r_jr_k$

• Forgetting the conjugate root when a real polynomial has a nonreal root.
• Applying Vieta's signs incorrectly, for instance writing $\sum r_k=a_{n-1}$ instead of $-a_{n-1}$.
• Expecting a real cubic with a complex root to have three real roots.

9.Trig identities through complex exponentials

Many trigonometric identities are binomial identities in disguise. Let $z=e^{i\theta}$. Then $\cos\theta=\frac{z+z^{-1}}{2}$ and $\sin\theta=\frac{z-z^{-1}}{2i}$, so any polynomial in $\cos\theta$ and $\sin\theta$ becomes a Laurent polynomial in $z$. The binomial theorem, orthogonality, and the symmetry $z\leftrightarrow z^{-1}$ then do the work, and the result is converted back to cosines and sines at the end.

Powers of cosine are the model case. Expanding $\cos^{n}\theta=\frac{1}{2^{n}}\left(z+z^{-1}\right)^{n}=\frac{1}{2^{n}}\sum_{k=0}^{n}\binom nk z^{\,n-2k}$ and grouping the terms $z^{m}$ and $z^{-m}$, which add to $2\cos m\theta$, gives $\cos^{n}\theta=\frac{1}{2^{n}}\sum_{k=0}^{n}\binom nk\cos\big((n-2k)\theta\big)$. For $n$ even the constant term appears once; for $n$ odd there is no constant term. For $n=3$ this gives $\cos^{3}\theta=\frac{3\cos\theta+\cos3\theta}{4}$.

Products of sines and cosines are handled the same way. From $\sin\alpha=\frac{e^{i\alpha}-e^{-i\alpha}}{2i}$ one gets $\sin\alpha\sin\beta=\frac{(e^{i\alpha}-e^{-i\alpha})(e^{i\beta}-e^{-i\beta})}{-4}$, and the four exponentials combine into $\frac{1}{2}\big[\cos(\alpha-\beta)-\cos(\alpha+\beta)\big]$. The same expansion gives every product-to-sum formula, so students need not memorize them separately.