Example 6. Find all complex numbers $z \neq 0$, satisfying the condition $z^{n-1}=\bar{z}$.
Solution. Let $z=\rho e^{i \varphi}$. Then $\bar{z}=\rho e^{-i \varphi}$.

According to the condition

$$
\rho^{n-1} e^{i(n-1) \varphi}=\rho e^{-i \varphi} \text { or } \rho^{n-2} e^{i n \varphi}=1,
$$

from which $\rho^{n-2}=1$, i.e., $\rho=1$, and $i \varphi=2 k \pi i$, i.e., $\varphi=\frac{2 k \pi}{n}(k=0,1,2, \ldots, n-1)$. Therefore,

$$
z_{k}=e^{i 2 \pi k / n}(k=0,1,2, \ldots, n-1)
$$

## Problems for Independent Solution

10. Represent the following complex numbers in trigonometric form:
a) -2 ;
b) $2 i$;
c) $-\sqrt{2}+i \sqrt{2}$;
d) $1-\sin \alpha+i \cos \alpha\left(0<\alpha<\frac{\pi}{2}\right)$
e) $\frac{1+\cos \alpha+i \sin \alpha}{1+\cos \alpha-i \sin \alpha} \quad\left(0<\alpha<\frac{\pi}{2}\right)$;

in exponential form:
f) -2 ;
g) $i$;
h) $-i$;
i) $-1-i \sqrt{3}$;
j) $\sin \alpha-i \cos \alpha\left(\frac{\pi}{2}<\alpha<\pi\right)$
k) $5+3 i$.

Let the complex numbers $z_{1}$ and $z_{2}$ be given in trigonometric form

$$
z_{1}=\rho_{1}\left(\cos \varphi_{1}+i \sin \varphi_{1}\right), \quad z_{2}=\rho_{2}\left(\cos \varphi_{2}+i \sin \varphi_{2}\right)
$$

Their product is found by the formula

$$
z_{1} z_{2}=\rho_{1} \rho_{2}\left[\cos \left(\varphi_{1}+\varphi_{2}\right)+i \sin \left(\varphi_{1}+\varphi_{2}\right)\right]
$$

i.e., when multiplying complex numbers, their moduli are multiplied, and their arguments are added:

$$
\left|z_{1} z_{2}\right|=\left|z_{1}\right| \cdot\left|z_{2}\right|, \quad \operatorname{Arg}\left(z_{1} z_{2}\right)=\operatorname{Arg} z_{1}+\operatorname{Arg} z_{2}
$$

The quotient of two complex numbers $z_{1}$ and $z_{2} \neq 0$ is found by the formula

$$
\frac{z_{1}}{z_{2}}=\frac{\rho_{1}}{\rho_{2}}\left[\cos \left(\varphi_{1}-\varphi_{2}\right)+i \sin \left(\varphi_{1}-\varphi_{2}\right)\right]
$$

i.e.

$$
\left|\frac{z_{1}}{z_{2}}\right|=\frac{\left|z_{1}\right|}{\left|z_{2}\right|}, \quad \operatorname{Arg} \frac{z_{1}}{z_{2}}=\operatorname{Arg} z_{1}-\operatorname{Arg} z_{2}
$$

Raising a complex number

$$
z=\rho(\cos \varphi+i \sin \varphi)
$$

to a natural power $n$ is done by the formula

$$
z^{n}=\rho^{n}(\cos n \varphi+i \sin n \varphi)
$$

i.e.

$$
\left|z^{n}\right|=|z|^{n}, \quad \operatorname{Arg} z^{n}=n \operatorname{Arg} z+2 \pi k \quad(k=0, \pm 1, \ldots)
$$

From this, we obtain de Moivre's formula

$$
\left(\cos \varphi+i \sin \varphi\right)^{n}=\cos n \varphi+i \sin n \varphi
$$

Properties of the modulus of complex numbers

1. $|\bar{z}|=|z|$
2. $z \bar{z}=|z|^{2}$
3. $\left|z_{1} z_{2}\right|=\left|z_{1}\right| \cdot\left|z_{2}\right|$
4. $\left|z^{n}\right|=|z|^{n}$;
5. $\left|\frac{z_{1}}{z_{2}}\right|=\frac{\left|z_{1}\right|}{\left|z_{2}\right|}, \quad z_{2} \neq 0$;
6. $|\operatorname{Re} z| \leqslant|z|, \quad|\operatorname{Im} z| \leqslant|z|$
7. $\left|z_{1}+z_{2}\right| \leqslant\left|z_{1}\right|+\left|z_{2}\right|$
8. $\left||z_{1}|-| z_{2}|\right| \leqslant\left|z_{1}-z_{2}\right|$.