Complex numbers

Solution 1.5.1

(See Exercise 1.4.) We have \(\frac{i-1}{i+1} = \frac{(i-1)\overline{i+1}}{|i+1|^2} = \frac{(i-1)(1-i)}{2} = i\). Thus \(z=i^3 = -i\) is the algebraic form. We have \(|z|=1\) and \({\mathrm {arg}} z = \frac 32 \pi\), so

\[ z = \cos(\frac 32 \pi) + i \sin \frac 32 \pi. \]

Solution 1.5.2

(See Exercise 1.5.) In order to solve

\[ z = 3i |z|\overline z \]

we would like to divide by \(z\). This is only possible if \(z \ne 0\), so we first consider the case \(z = 0\). In this case both sides of the equation are equal to 0, so \(z=0\) is indeed a solution. Now, we consider \(z \ne 0\) and divide the above equation by \(z\) and obtain

\[ 1 = 3i |z| \frac{\overline z}z. \]

There are different ways to solve this equation. One may put \(z = a+ib\) and solve the resulting quadratic equation. For illustrational purposes, we rather consider the trigonometric form \(z = r (\cos \alpha + i \sin \alpha)\). Then \(\overline z = r (\cos \alpha - i \sin \alpha) = r (\cos (-\alpha) + i \sin (-\alpha))\), and

\[ |z| \frac{\overline z}z = r \frac{r (\cos (-\alpha) + i \sin (-\alpha))}{r (\cos (\alpha) + i \sin (\alpha))}. \]

Note that \(r = |z| \ne 0\), so we can cancel this in the right-hand fraction. We have

\[ \frac{\cos (-\alpha) + i \sin (-\alpha)}{\cos (\alpha) + i \sin (\alpha)} = \cos(-2\alpha) + i \sin(-2\alpha). \]

We then obtain

\[ \frac 1{3i} = -\frac 13 i = |z| \frac{\overline z}z = r (\cos(-2\alpha) + i \sin(-2\alpha)). \]

This implies \(r = \frac 13\). Concerning the arguments, we have to be more careful: the above equation is equivalent to saying that

\[ -2 \alpha \equiv \frac 32 \pi \mod 2 \pi \]

cf. around Equation (1.6). There are two solutions: \(-2 \alpha = \frac 32 \pi\) or \(- 2\alpha = \frac 32 \pi + 2 \pi\). The former yields \(\alpha = -\frac 34 \pi\), the latter \(\alpha = \frac \pi 4\). (Of course, we can now add integer multiples of \(2\pi\) to these values of \(\alpha\), so \(\alpha = \frac 54 \pi\) is another solution. However, this gives the same value for \(z\).) The resulting solutions are

\[ z = \frac 1 3 (\cos(-\frac 34 \pi) + i \sin (-\frac 34 \pi)) \]

and

\[ z = \frac 1 3 (\cos(\frac 14 \pi) + i \sin (\frac 14 \pi)). \]

To sum up, the above equation has three solutions, \(z = 0\) and these two solutions.

Solution 1.5.3

(See Exercise 1.6.) There are three solutions, namely

\[ z_0 = \sqrt 3 - i = 2 (\cos (-\pi/6) + i \sin(-\pi/6)), \]

\[ z_1 = -\sqrt 3 - i = 2 (\cos (-5\pi/6) + i \sin(-5\pi/6)), \]

\[ z_2 = 2i = 2 (\cos (-3\pi/2) + i \sin(-3\pi/2)). \]

Here is a picture: