## Task B-4.7.

Determine all complex numbers $z$ such that $\operatorname{Re} z>0$ and $z^{8}+(1-4 i) z^{4}-4 i=0$.
## Solution.

By introducing the substitution $z^{4}=t$, we obtain the quadratic equation

$$
t^{2}+(1-4 i) t-4 i=0
$$

Solving equation $(*)$ yields

$$
t_{1,2}=\frac{-1+4 i \pm \sqrt{(1-4 i)^{2}+16 i}}{2}=\frac{-1+4 i \pm \sqrt{(1+4 i)^{2}}}{2}=\frac{-1+4 i \pm(1+4 i)}{2}
$$

i.e., $t_{1}=4 i, t_{2}=-1$.

If we write $t_{1}=4 i$ in trigonometric form as $t_{1}=4\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$, from $z^{4}=t_{1}$ we get

$$
z_{1, k}=\sqrt{2}\left(\cos \frac{(4 k+1) \pi}{8}+i \sin \frac{(4 k+1) \pi}{8}\right), \quad k=0,1,2,3
$$

From (**) it follows:

for $k=0, z_{1,0}=\sqrt{2}\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right)$,

for $k=1, z_{1,1}=\sqrt{2}\left(\cos \frac{5 \pi}{8}+i \sin \frac{5 \pi}{8}\right)$,

for $k=2, z_{1,2}=\sqrt{2}\left(\cos \frac{9 \pi}{8}+i \sin \frac{9 \pi}{8}\right)$,

for $k=3, z_{1,3}=\sqrt{2}\left(\cos \frac{13 \pi}{8}+i \sin \frac{13 \pi}{8}\right)$.

From the condition $\operatorname{Re} z>0$, it follows that the solutions are $z_{1,0}$ and $z_{1,3}$.

2 points

Similarly, if we write $t_{2}=-1$ in trigonometric form as $t_{2}=\cos \pi+i \sin \pi$, from $z^{4}=t_{2}$ we get

$$
z_{2, k}=\cos \frac{(2 k+1) \pi}{4}+i \sin \frac{(2 k+1) \pi}{4}, \quad k=0,1,2,3
$$

From (***) it follows:

for $k=0, z_{2,0}=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}$,

for $k=1, z_{2,1}=\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}$,

for $k=2, z_{2,2}=\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}$,

for $k=3, z_{2,3}=\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}$.

From the condition $\operatorname{Re} z>0$, it follows that the solutions are $z_{2,0}$ and $z_{2,3}$.

2 points

Therefore, the solutions are:

$z \in\left\{\sqrt{2}\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right), \sqrt{2}\left(\cos \frac{13 \pi}{8}+i \sin \frac{13 \pi}{8}\right), \cos \frac{\pi}{4}+i \sin \frac{\pi}{4}, \cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\right\}$.