# Complex numbers

### Trigonometric values via complex numbers

B5, 2012

Using the steps below, find the value of $$x^{2012}+x^{-2012},$$ where $$x+x^{-1}=\frac{\sqrt{5}+1}{2}$$

(a) For any real $$r,$$ show that $$\left|r+r^{-1}\right| \geq 2 .$$ What does this tell you about the given $$x ?$$

(b) Show that $$\cos \left(\frac{\pi}{5}\right)=\frac{\sqrt{5}+1}{4}$$. Compare $$\sin \left(\frac{2 \pi}{5}\right)$$ and $$\sin \left(\frac{3 \pi}{5}\right)$$

(c) Combine conclusions of parts a and b to express $$x$$ and therefore the desired quantity in a suitable form.

Solution

(a) Because of the absolute value we may assume that $$r > 0$$ by replacing $$r$$ with $$-r$$ if necessary. Now use AM-GM inequality or the fact that $$(\sqrt{r}-\sqrt{1 / r})^{2} \geq 0 .$$ since $$x+x^{-1}=\frac{\sqrt{5}+1}{2}<2$$ given $$x$$ must be a non-real complex number.

(b) Let $$\theta=\frac{\pi}{5} .$$ Then $$\sin (2 \theta)=\sin (\pi-2 \theta)=\sin (3 \theta) .$$ Using the formulas for $$\sin (2 \theta)$$ and $$\sin (3 \theta),$$ canceling $$\sin \theta\left(\text { it is nonzero) and substituting } \sin^{2} \theta=1-\cos ^{2} \theta,\right.$$ gives the quadratic equation $$4 \cos ^{2} \theta-2 \cos \theta-1=0 .$$ since $$\cos \theta > 0$$, we get $$\cos \theta=\frac{\sqrt{5}+1}{4}$$

(c) Let $$x=d e^{i \alpha}=d(\cos \alpha+i \sin \alpha) .$$ Then $$x^{-1}=d^{-1} e^{-i \alpha}=d^{-1}(\cos \alpha-i \sin \alpha) .$$ Adding and using that $$x+x^{-1}=\frac{\sqrt{5}+1}{2}=2 \cos \left(\frac{\pi}{5}\right),$$ we get $$d=1$$ and $$\alpha=\pm \theta .$$ So $$x=e^{\pm \frac{1 \pi}{5}}$$.
\begin{align} x^{2012}+x^{-2012} &= 2 \cos \left(\frac{2012 \pi}{5}\right)\\ &=2 \cos \left(402 \pi+\frac{2 \pi}{5}\right)\\ &=2 \cos \left(\frac{2 \pi}{5}\right)\\ &=2 \cos ^{2}\left(\frac{\pi}{5}\right)-1 \\ &=\frac{\sqrt{5}-1}{2}\\ \end{align}

### Set of powers

A5, 2021

For a complex number $$z$$ let $$P(z): \{z^k: k \in \{1,2,3,\ldots\} \}$$.
(a) For every natural number $$n$$ there exits a $$z$$ such that $$P(z)=n$$.
(b) There exits a unique $$z$$ such that $$P(z)=3$$.
(c) If $$|z|\neq 1$$ and $$|z|\neq 0$$ then $$P(z)$$ is an infinite set.
(d) $$P(e^i)$$ has infinite elements.

Solution (a) True. Pick $$z=e^{\frac{2\pi i}{n}}$$.
(b) False. Both $$\omega$$ and $$\omega^2$$ work.
(c) True. If $$z=re^{i\theta}$$, then every number in $$P$$ has a different length.
(d) True. If $$e^{ia}=e^{ib}$$ for some powers $$a,b$$ with $$a> b$$, then $$a=2k\pi+b$$ for some natural number $$k$$. But this would imply that $$\pi$$ is rational.

### Complex polygon

B2, 2020

2. (i) Let $$z=e^{\frac{2i\pi}{n}}$$ where $$n\geq 2$$ is a positive integer. Prove that $$\sum_{k=0}^{n-1}z^k=0.$$

Solution

Since $$z^n=1$$, we have $$z^n-1=0$$.

$z^n-1 = (z^{n-1} + z^{n-2} + \cdots + 1)(z-1) = 0$

For $$n\geq 2$$, $$z\neq 1$$. So the first factor must be zero. This proves the statement.

2. (ii) Prove that $$\cos 1^\circ + \cos 41^\circ + \cos 81^\circ + \cdots + \cos 321^\circ = 0$$

Solution

\begin{align} A &:= \cos 1^\circ + \cos 41^\circ + \cos 81^\circ + \cdots + \cos 321^\circ \\ B &:= \sin 1^\circ + \sin 41^\circ + \sin 81^\circ + \cdots + \sin 321^\circ \\ \end{align}

Notice that $$40^\circ=2\pi/9$$. Let $$\theta = 1^\circ = \pi/180$$. Then:

$A+iB = e^{i\theta} \left( \sum_{k=0}^{8} e^{ \frac{2\pi i}{9}k } \right)$

From problem 2(i), we know that RHS of the above equation is zero. Since $$A$$ and $$B$$ are real numbers, both of them must be individually zero. In particular, $$A=0$$, which proves the statement.

### Power of a complex number

A13, 2010

If $$b$$ is a real number satisfying $$b^{4}+\frac{1}{b^{4}}=6,$$ find the value of $$\left(b+\frac{i}{b}\right)^{16}$$ where $$i=\sqrt{-1}$$

Solution

\begin{align} \left(b^{2}\right)^{2}+\left(\frac{i^{2}}{b^{2}}\right)^{2}&=6\\ \left(b^{2}+\frac{i^{2}}{b^{2}}\right)^{2}&=4\\ b^{2}+\frac{i^{2}}{b^{2}}&=\pm 2 \end{align}

Let us know look at the quantity we want to compute:

\begin{align} \left(b+\frac{i}{b}\right)^{16}&=\left(b^{2}+\frac{i^{2}}{b^{2}}+2 i\right)^{8}\\ &=(\pm 2+2i)^{8}\\ &=2^{8}\left(\sqrt{2} e^{\frac{i \pi}{4}}\right)^{8} \text{ or } 2^{8}\left(\sqrt{2} e^{\frac{3i \pi}{4}}\right)^{8} \\ &=2^{12}\\ &=4096 \end{align}

### Complex triangle

A7, 2013

Let $$A, B, C$$ be angles such that $$e^{i A}, e^{i B}, e^{i C}$$ form an equilateral triangle in the complex plane. Find the values of the given expressions.

a) $$e^{i A}+e^{i B}+e^{i C}$$

b) $$\cos A+\cos B+\cos C$$

c) $$\cos 2 A+\cos 2 B+\cos 2 C$$

d) $$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C$$

Solution

a) $$e^{i A}+e^{i B}+e^{i C}=0$$ by taking the vector sum of the three points on the unit circle.

b) $$\cos A+\cos B+\cos C=0=$$ real part of $$e^{i A}+e^{i B}+e^{i C},$$ which is 0 by part a.

c) $$\cos 2 A+\cos 2 B+\cos 2 C=0$$ because the points $$e^{2 i A}, e^{2 i B}, e^{2 i C}$$ on the unit circle also form an equilateral triangle in the complex plane, since taking $$B=A+(2 \pi / 3), C=A+(4 \pi / 3)$$, we get $$2 B=2 A+(4 \pi / 3)$$ and $$2 C=2 A+(8 \pi / 3)=2 A+(2 \pi / 3)+2 \pi$$. The last term $$2 \pi$$ does not change the position of the point.

d) $$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=\frac{3}{2}$$ because, using the formula for $$\cos 2 \theta$$ in part $$c,$$ we get $$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=\sin ^{2} A+\sin ^{2} B+\sin ^{2} C$$ and the sum of the LHS and the RHS in this equation is $$3 .$$

### Maximum and minimum of an average

A9, 2014

Let $$\theta_{1}, \theta_{2}, \ldots, \theta_{13}$$ be real numbers and let $$A$$ be the average of the complex numbers $$e^{i \theta_{1}}, e^{i \theta_{2}} \ldots, e^{i \theta_{13}},$$ where $$i=\sqrt{-1}$$. As the values of $$\theta$$ 's vary over all 13 -tuples of real numbers, find

(i) the maximum value attained by $$|A|$$

(ii) the minimum value attained by $$|A|$$.

Solution

(i) Each $$e^{i\theta}$$ can take a maximum value of 1, which is attained when $$\theta=0$$. Hence, the maximum average is also 1.

(ii) To get the minimum, place the 13 points on the vertices of a regular cyclic polygon. The average corresponds to the center of the polygon which is $$(0,0)$$. Hence, the minimum value $$|A|$$ can take is 0.

### Roots of unity I

A10, 2015

(i) Suppose $$z_{1}, z_{2}$$ are complex numbers. One of them is in the second quadrant and the other is in the third quadrant. How does $$||z_{1}|-| z_{2}||$$ compare with $$\left|z_{1}+z_{2}\right|$$?

(ii) Complex numbers $$z_{1}, z_{2}$$ and 0 form an equilateral triangle. How does $$\left|z_{1}^{2}+z_{2}^{2}\right|$$ compare with$$\left|z_{1} z_{2}\right|$$.

(iii) Let $$1, z_{1}, z_{2}, z_{3}, z_{4}, z_{5}, z_{6}, z_{7}$$ be the complex 8 -th roots of unity. Find the value of $$\prod_{i=1, \ldots, 7}^{\Pi}\left(1-z_{i}\right),$$ where the symbol $$\Pi$$ denotes product.

Solution

(i) ||$$z_{1}|-| z_{2}||<\left|z_{1}+z_{2}\right| .$$ One way: using triangle inequality for $$z_{1}+z_{2}$$ and $$-z_{2}$$ we get $$\left|z_{1}\right| \leq\left|z_{1}+z_{2}\right|+\left|-z_{2}\right|$$ and so $$\left|z_{1}\right|-\left|z_{2}\right| \leq\left|z_{1}+z_{2}\right| .$$ Now we may take absolute value on the LHS because switching $$z_{1}$$ and $$z_{2}$$ keeps RHS the same. For equality, $$z_{1}+z_{2}$$ and $$-z_{2}$$ must point in the same direction, so $$z_{1}$$ and $$z_{2}$$ must be along the same line. But they are in quadrants 2 and $$3,$$ so this cannot happen.

(ii) $$z_{2}$$ must be obtained by rotating $$z_{1}$$ by angle $$\pi / 3,$$ say in the counterclockwise direction (otherwise interchange the two). Then $$z_{2}=z_{1} e^{\frac{\pi i}{3}}$$. Then $$z_{1}^{2}+z_{2}^{2}=z_{1}^{2}\left(1+e^{\frac{2 \pi i}{3}}\right)$$ and $$z_{1} z_{2}=z_{1}^{2} e^{\frac{\pi i}{3}} . \quad$$ Now $$1+e^{\frac{2 \pi i}{3}}=e^{\frac{\pi i}{3}}$$ (see by calculation or picture), so we have in fact $$z_{1}^{2}+z_{2}^{2}=z_{1} z_{2}$$

(iii) We have $$\prod_{i=1, \ldots, 7}\left(x-z_{i}\right)=\frac{x^{8}-1}{x-1}=1+x+\ldots+x^{7}$$. Putting $$x=1$$ gives $$\prod_{1=1, \ldots, 7}\left(1-z_{i}\right)=8$$

### Counting roots in a quadrant

A6, 2018

Consider the equation

$z^{2018}=2018^{2018}+i$

where $$i=\sqrt{-1}$$

(a) How many complex solutions does this equation have?

(b) How many solutions does each of the four quadrants have?

Solution

(a) In general, the equation $$z^n = re^{i\theta}$$ has $$n$$ solutions given by: $r^{1 / n} \exp \left[\frac{i(\theta+2 k \pi)}{n}\right] \text{ for each } 0 \leq k \leq n-1$ The given equation has 2018 complex solutions, since we can express the complex number in the RHS as $$re^{i\theta}$$ for some small $$\theta$$.

(b) Instead of looking at the given equation, first consider the solutions to: $$x^{2018}=r$$. Two of them are real values: $$r^{1/2018}$$ and $$-(r)^{1/2018}$$. The other 2016 are distributed equally in the four quadrants, 504 each.

If we rotate the solutions to $$x^{2018}=r$$ by a tiny angle in the counter-clockwise direction, we get the solutions to the given equation. (The value of $$r$$ being $$\sqrt{ 2018^{2018\cdot 2} + 1^2 }$$). This gives 505 values each in the first and third quadrant but still 504 in the second and fourth quadrant.

### Counting the roots outside a disk

B2A, 2019

How many complex roots $$w$$ of the equation $$z^{2019}-1=0$$ satisfy $$|w+1| \geq \sqrt{2+\sqrt{2}}$$

Solution

Such roots can be expressed as follows

$w=\frac{\cos (2 \pi k)}{2019}+i \frac{\sin (2 \pi k)}{2019} \quad \text { for } k=0,\pm 1, \ldots,\pm 1009$

Therefore,

$|w+1|^{2}=2+2 \frac{\cos (2 \pi k)}{2019}$

Hence we want to find all $$k$$ such that

$\frac{\cos (2 \pi k)}{2019} \geq \frac{1}{\sqrt{2}}$

Which is same as

$\begin{array}{l} \quad\left|\frac{2 \pi k}{2019}\right| \leq \frac{\pi}{4} \\ \text { i.e. }|k| \leq 252 \end{array}$

So there are 505 solutions.