Example 6. Write in algebraic form Arctg $(1+i)$.

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Solution. Substituting $z=1+i$ into formula (9), we get

$$
\operatorname{Arctg}(1+i)=-\frac{i}{2} \operatorname{Ln} \frac{1+i(1+i)}{1-i(1+i)}=-\frac{i}{2} \operatorname{Ln} \frac{i}{2-i}=-\frac{i}{2} \operatorname{Ln}\left(-\frac{1}{5}+\frac{2}{5} i\right)
$$

Further,

$$
\operatorname{Ln}\left(-\frac{1}{5}+\frac{2}{5} i\right)=-\ln \sqrt{5}+(2 k+1) \pi i-i \operatorname{arctg} 2
$$

Finally,

$$
\operatorname{Arctg}(1+i)=-\frac{1}{2} \operatorname{arctg} 2+(2 k+1) \frac{\pi}{2}+\frac{i}{2} \ln \sqrt{5} \quad(k=0, \pm 1, \pm 2, \ldots)
$$

## Problems for Independent Solution

Write the following complex numbers in algebraic form:
69. a) $e^{i \pi / 4}$;
b) $\ln (1-i)$.
70. a) $\sin \pi i$;
b) $\cos \pi i$
c) $\operatorname{tg} \frac{\pi}{2} i$.

71. a) $\operatorname{ctg} \pi i$;
b) $\operatorname{Arcsin} i$;
c) $\operatorname{Arctg} \frac{i}{3}$.
72. a) $\operatorname{Arccos} i$;

b) $\operatorname{sh} \frac{\pi i}{2}$;

c) th $\pi i$.