## 10.5 Find:

a) the real part of the complex number $z=\frac{(2-i)^{3}}{3+4 i}$;

b) the imaginary part of the complex number $z=\left(\frac{1+i}{1-i}\right)^{11}$.

Perform the operations (10.6-10.8):
10.5 a) We have

$$
\begin{aligned}
& \frac{(2-i)^{3}}{3+4 i}=\frac{(2-i)^{3}(3-4 i)}{(3+4 i)(3-4 i)}=\frac{\left(8-3 \cdot 4 i+3 \cdot 2 i^{2}-i^{3}\right)(3-4 i)}{25}= \\
& =\frac{(2-11 i)(3-4 i)}{25}=\frac{-38-41 i}{25}=-1.52-1.64 i
\end{aligned}
$$

i.e., $\operatorname{Re} z=-1.52$.

b) Since

$\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{(1+i)^{2}}{2}=\frac{2 i}{2}=i$,

then $z=i^{11}=i^{3}=-i$. Therefore, $\operatorname{Im} z=-1$.

Answer: a) $-1.52$; b) $-1$.