## Multiplication and division of complex numbers in trigonometric form

If $z_1 = r_1 ( \cos \varphi_1 + i \sin \varphi_1)$ and $z_2 = r_2 ( \cos \varphi_2 + i \sin \varphi_2)$, $z_2 \neq 0$, then their **product** is defined as

**$$z_1 \cdot z_2 = r_1 r_2 ( \cos (\varphi_1 + \varphi_2) + i \sin (\varphi_1 + \varphi_2)),$$**

and their **quotient** as

**$$\frac{z_1}{z_2} = \frac{r_1}{r_2} ( \cos (\varphi_1 – \varphi_2) + i \sin (\varphi_1 – \varphi_2)).$$**

* Example 1:* Calculate $z_{1}\cdot z_{2}$, where $z_{1} = 2 \cdot \left(cos \frac{\pi}{5} + isin \frac{\pi}{5}\right)$ and $z_{2} = 5 \cdot \left(cos \frac{2\pi}{3} + isin \frac{2\pi}{3}\right)$.

**Solution:**

$z_1 \cdot z_2 = 2 \cdot 5 \cdot \left(cos \left(\frac{\pi}{5}+\frac{2\pi}{3}\right) + isin \left(\frac{\pi}{5}+\frac{2\pi}{3}\right)\right)$

$= 10 \cdot \left(cos \frac{13 \pi}{15} + isin \frac{13 \pi}{15}\right)$

* Example 2:* Calculate $\frac{z_{1}}{z_{2}}$, where $z_{1} = 3 \cdot \left(cos \frac{\pi}{3} + isin \frac{\pi}{3}\right) $ and $z_{2} = 4 \cdot \left(cos \frac{\pi}{6} + isin \frac{\pi}{6}\right)$.

**Solution:**

$\frac{z_{1}}{z_{2}} = \frac{3}{4} \cdot \left(cos \left(\frac{\pi}{3} – \frac{\pi}{6}\right) + isin \left(\frac{\pi}{3} – \frac{\pi}{6}\right)\right)$

$ = \frac{3}{4} \cdot \left(cos \left(\frac{\pi}{6}\right) + isin \left( \frac{\pi}{6}\right)\right)$

**Exponentiation of complex numbers. De Moivre’s formula**

For every complex number $ z = r (\cos \varphi + i \sin \varphi)$ and $n \in \mathbb{N}$,

**$$z^n = r^n ( \cos n \varphi + i \sin n \varphi)$$**

is valid.

The formula above is called the **De Moivre’s formula**.

* Example 3: *Calculate

$$ \frac{(-1 + i)^{20}}{(\sqrt{2} – i \sqrt{2})^{10}}.$$

**Solution**:

Let $z_1 = -1 + i $ and $z_2 = \sqrt{2} – i \sqrt{2}$.

We first need to find the trigonometric form of $z_1$:

$$r_1 = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$$

$$ tg \varphi_1 = -1 \Rightarrow \varphi_1 = \frac{3\pi}{4},$$

because $z_1$ is in the second quadrant. Therefore

$$z_1 = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right).$$

Now, by using the De Moivre’s formula, we have:

$$z_1^{20} = (-1 + i)^{20} $$

$$=(\sqrt{2})^{20} \left ( \cos \left(20\cdot \frac{3\pi}{4}\right) + i \sin \left(20\cdot \frac{3\pi}{4}\right)\right) $$

$$= 1024 ( \cos (15\pi) + i \sin (15\pi)$$

$$= 1024 (\cos \pi + i \sin \pi) $$

For $z_2$ we have:

$$r_2 = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2+2} = \sqrt{4} = 2$$

$$tg \varphi_2 = -1 \Rightarrow \varphi_2 = \frac{7\pi}{4},$$

because the number $z_2$ is located in the fourth quadrant. It follows that the trigonometric form of the complex number $z_2$ is:

$$z_2 = 2 \left( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}\right).$$

Using the De Moivre’s formula, we have:

$$z_2^{10} = (\sqrt{2} – i \sqrt{2})^{10}= $$

$$= 2^{10} \left ( \cos \left(10\cdot \frac{7\pi}{4}\right) + i \sin \left(10\cdot \frac{7\pi}{4}\right)\right)$$

$$ = 1024 \left ( \cos \left(\frac{35 \pi}{2}\right) + i \sin \left(\frac{35 \pi}{2}\right) \right) $$

$$= 1024 \left ( \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) \right) $$

Now we just need to calculate the quotient:

$$\frac{z_1^{20}}{z_2^{10}} $$

$$= \frac{(-1 + i)^{20}}{(\sqrt{2} – i \sqrt{2})^{10}} $$

$$=\frac{1024}{1024 } \cdot \left ( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) $$

$$ = 1 \cdot \left ( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) $$

$$ = i$$

**$n$th root of complex numbers**

Every complex number $z= r (\cos \varphi + i \sin \varphi)$ has exactly $n$ different values of $n$-th root given by the formula:

**$$\sqrt[n]{z} = \sqrt[n]{r} \left ( \cos \frac{\varphi + 2 k \pi}{n} + i \sin \frac{\varphi + 2 k \pi}{n} \right),$$**

where $k \in \{0, 1, 2, …, n-1\}$ and $n \in \mathbb{N}$.

* Example 4:* Determine $\sqrt[6]{1}$.

**Solution**:

The number $z=1 + 0\cdot i$ has the following trigonometric form:

$$ r = \sqrt{1^2 + 0^2} = \sqrt{1} = 1$$

$$ tg \varphi = 0 \Rightarrow \varphi = 0$$

$$\Rightarrow z = 1 (\cos 0 + i \sin 0)$$

$$z = 1.$$

By using the formula for the $n$th root of a complex number for $n=6$, we have

$$\sqrt[6]{1} = \sqrt[6]{1} \left ( \cos \frac{0 + 2k\pi}{6} + i \sin \frac{0 + 2 k \pi}{6} \right).$$

For $k=0$:

$$z_0 = 1 (\cos 0 + i \sin 0) = 1.$$

For $k = 1$:

$$z_1 = 1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = \frac{1}{2} + i \frac{\sqrt{3}}{2}.$$

For $k=2$:

$$z_2 = 1 \left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right) = – \frac{1}{2} + i \frac{\sqrt{3}}{2}.$$

For $k=3$:

$$ z_3 = 1 (\cos \pi + i \sin \pi) = -1.$$

For $k=4$:

$$z_4 = 1 \left ( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right ) = – \frac{1}{2} – i \frac{\sqrt{3}}{2}.$$

For $k=5$:

$$z_5 = 1 \left ( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) = \frac{1}{2} – i \frac{\sqrt{3}}{2}.$$

We can notice that obtained complex numbers are arranged symmetrical. They form the vertices of a regular hexagon.

Similar holds true in general. The formula for the $n$th root of complex numbers has a geometric interpretation. All $n$th roots of complex numbers have the same modulus $\sqrt[n]{r}$, which means that they are equally distanced from the origin. Therefore, they all lie on the **circle with center at the origin and radius $\sqrt[n]{r}$**.

Furthermore, arguments of every two consecutive $n$th roots differ for $ \frac{2 \pi}{n}$, which means that these **$n$th roots are arranged so that they form the vertices of a regular polygon**.