10.47 Given complex numbers $z_{1}=1+a i$ and $z_{2}=2^{\frac{3}{4}}\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right)$. Find all real values of $a$ for which $z_{1}^{3}=z_{2}^{2}$.
10.47 We find

\[
\begin{aligned}
& z_{1}^{3}=(1+a i)^{3}=\left(1-3 a^{2}\right)+\left(3 a-a^{3}\right) i \\
& z_{2}^{2}=\left(2^{\frac{3}{4}}\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right)\right)^{2}= \\
& =2^{\frac{3}{2}}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)=2^{\frac{3}{2}}\left(-\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}\right)=-2+2 i
\end{aligned}
\]

Since the numbers $z_{1}^{3}$ and $z_{2}^{2}$ are equal, we arrive at the system

\[
\left\{\begin{array}{l}
1-3 a^{2}=-2 \\
3 a-a^{3}=2
\end{array}\right.
\]

from which, considering that $a \in \boldsymbol{R}$, we obtain $a=1$.