complex analysis conjugate problem

Question

Is it equal? Or is it not equal? Also I tried two way. Logarithmic and polar representation. But I failed. Sorry for my English. I couldn't show this.

$$\overline{(z^{z})} \neq \bar{z}^{\bar{z}}$$ $$\overline{(z^{z})} = \bar{z}^{\bar{z}}$$

Answer 1

Take the case $z=i$. Then $$z^z=i^i=e^{i\log{i}}=e^{i(i2n\pi+i\pi/2)}=e^{-2n\pi-\pi/2}$$ Since this is real, for all integer values of $n$, we have $\overline{z^z}=z^z$ in this case. On the other hand, $$\overline{z}^{\overline{z}}=(-i)^{-i}=e^{-i(i2k\pi-i\pi/2)}=e^{-2k\pi+\pi/2}$$ There are no integer values of $k$ and $n$ such that $-2n\pi-\pi/2=-2k\pi+\pi/2$, so in the case $z=i$, there is no value of the left-hand side equal to any value of the right-hand side.

Therefore, as a general rule, $\overline{z^z}\neq\overline{z}^{\overline{z}}$. Of course, they're equal when $z$ is real.