3.3. Calculate: a) $(1+i)^{11}$, b) $(1-i \sqrt{3})^{73}$.
Solution. Consider part a). Let $1+i=z_{0}$, then, by the condition, it is necessary to raise the number $z_{0}$ to the 11th power, which is most conveniently done if the number $z_{0}$ is represented in trigonometric or exponential form. Choose the interval of variation of the principal value $\operatorname{Arg} z$, i.e., $\arg z$, as the interval $[0,2 \pi)$, then, applying the previously shown algorithm, we get:

$$
z_{0}=1+i=\sqrt{2}\left(\cos \left(\frac{\pi}{4}+2 \pi k\right)+i \sin \left(\frac{\pi}{4}+2 \pi k\right)\right)=\sqrt{2} e^{i\left(\frac{\pi}{4}+2 \pi k\right)}, k \in \mathbb{Z}
$$

Further, taking into account the formula of de Moivre, we have:

$$
\begin{aligned}
z_{0}^{11}= & \left.(1+i)^{11}=\left(\sqrt{2}\left(\cos \left(\frac{\pi}{4}+2 \pi k\right)+i \sin \left(\frac{\pi}{4}+2 \pi k\right)\right)\right)\right)^{11}= \\
& =2^{\frac{11}{2}}\left(\cos \left(\frac{11 \pi}{4}+22 \pi k\right)+i \sin \left(\frac{11 \pi}{4}+22 \pi k\right)\right)= \\
= & 2^{\frac{11}{2}}\left(\cos \left(\frac{3 \pi}{4}+22 \pi k\right)+i \sin \left(\frac{3 \pi}{4}+22 \pi k\right)\right), k \in \mathbb{Z}
\end{aligned}
$$

Or if we convert to algebraic form:

$$
z_{0}^{11}=2^{\frac{11}{2}}\left(-\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}\right)=2^{5}(-1+i)=-32+32 i
$$

Performing the same in part b), only representing the number $z_{1}=1-i \sqrt{3}$ in exponential form and choosing the interval of variation of the principal value $\operatorname{Arg} z$, i.e., $\arg z$, as the interval $[-\pi, \pi)$, we get:

$$
z_{1}^{73}=\left(2 e^{\left(\frac{\pi}{3}+2 \pi k\right)}\right)^{73}=2^{73} e^{i\left(-\frac{73 \pi}{3}+146 \pi k\right)}=2^{73} e^{i\left(-\frac{\pi}{3}+146 \pi k\right)}
$$

Or if we convert to algebraic form:

$$
\begin{aligned}
z_{1}^{73}= & 2^{73}\left(\cos \left(-\frac{\pi}{3}+146 \pi k\right)+i \sin \left(-\frac{\pi}{3}+146 \pi k\right)\right)= \\
& =2^{73}\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=2^{72}-2^{72} \sqrt{3} i=2^{72} z_{0}
\end{aligned}
$$

Note. One can act differently, for example, if $n \in N$ and $z=x+i y$, then, applying the binomial formula of Newton to the number $z^{n}$, we get:

$$
z^{n}=(x+i y)^{n}=\sum_{k=0}^{n} C_{n}^{k} x^{n-k}(i y)^{k}=\sum_{k=0}^{n} i^{k} C_{n}^{k} x^{n-k} y^{k}, \quad C_{n}^{k}=\frac{n!}{k!(n-k)!}
$$