3.4. Calculate: a) $\sqrt[3]{1+i}$, b) $\sqrt[4]{-1-i \sqrt{3}}$.
Solution. Consider part a). Let $1+i=z_{0}$, then, by the condition, it is necessary to find the cube root of the number $z_{0}$. The function of a complex variable $f(z)=\sqrt[n]{z}, n \in \mathbb{R}$ is multivalued and in any domain $D$ of the complex plane, not containing contours that encircle the point $z=0$, allows the selection of $n$ single-valued branches. Given this, since the point $z_{0}$ can be placed in such a domain, we will find all 3 values of $\sqrt[3]{z_{0}}$ (up to $\operatorname{Arg} z$). This is conveniently done if the number $z_{0}$ is represented, as above, in trigonometric or exponential form. We will choose the interval of variation of the principal value of $\operatorname{Arg} z$, i.e., $\arg z$, as the interval $[0,2 \pi)$ and, using the results of part a) of example 3.3, we get:

$$
\begin{gathered}
\sqrt[3]{z_{0}}=\sqrt[3]{1+i}=\sqrt[3]{\sqrt{2}\left(\cos \left(\frac{\pi}{4}+2 \pi k\right)+i \sin \left(\frac{\pi}{4}+2 \pi k\right)\right)}= \\
=2^{\frac{1}{6}}\left(\cos \left(\frac{\pi}{12}+\frac{2}{3} \pi k+2 \pi l\right)+i \cos \left(\frac{\pi}{12}+\frac{2}{3} \pi k+2 \pi l\right)\right) \\
k=0,1,2, \quad l \in \mathbb{Z}
\end{gathered}
$$

Thus, we have:

$$
\sqrt[3]{z_{0}}=\left[\begin{array}{l}
2^{\frac{1}{6}}\left(\cos \left(\frac{\pi}{12}+2 \pi l\right)+i \cos \left(\frac{\pi}{12}+2 \pi l\right)\right), k=0, \quad l \in \mathbb{Z} \\
2^{\frac{1}{6}}\left(\cos \left(\frac{3 \pi}{4}+2 \pi l\right)+i \cos \left(\frac{3 \pi}{4}+2 \pi l\right)\right), \quad k=1, \quad l \in \%^{\prime \prime} \\
2^{\frac{1}{6}}\left(\cos \left(\frac{17 \pi}{12}+2 \pi l\right)+i \cos \left(\frac{17 \pi}{12}+2 \pi l\right)\right), k=2, \quad l \in \text { \%. }^{\prime \prime}
\end{array}\right.
$$

Performing the same in part b), we get:

$$
\begin{gathered}
\sqrt[4]{-1-i \sqrt{3}}=\sqrt[4]{2\left(\cos \left(\frac{4 \pi}{3}+2 \pi k\right)+i \sin \left(\frac{4 \pi}{3}+2 \pi k\right)\right)}= \\
=\sqrt[4]{2}\left(\cos \left(\frac{\pi}{3}+\frac{1}{2} \pi k+2 \pi l\right)+i \sin \left(\frac{\pi}{3}+\frac{1}{2} \pi k+2 \pi l\right)\right) \\
k=0,1,2,3, \quad l \in \mathbb{Z}
\end{gathered}
$$

Or, equivalently:

$$
\sqrt[4]{-1-i \sqrt{3}}=\left[\begin{array}{l}
\sqrt[4]{2}\left(\cos \left(\frac{\pi}{3}+2 \pi l\right)+i \sin \left(\frac{\pi}{3}+2 \pi l\right)\right), k=0, \quad l \in \mathbb{Z} \\
\sqrt[4]{2}\left(\cos \left(\frac{5 \pi}{6}+2 \pi l\right)+i \sin \left(\frac{5 \pi}{6}+2 \pi l\right)\right), k=1, \quad l \in \mathbb{Z} \\
\sqrt[4]{2}\left(\cos \left(\frac{4 \pi}{3}+2 \pi l\right)+i \sin \left(\frac{4 \pi}{3}+2 \pi l\right)\right), k=2, \quad l \in \mathbb{Z} \\
\sqrt[4]{2}\left(\cos \left(\frac{11 \pi}{6}+2 \pi l\right)+i \sin \left(\frac{11 \pi}{6}+2 \pi l\right)\right), k=3, \quad l \in \mathbb{Z}
\end{array}\right.
$$