= Solution
Away from the negative real axis, the <complex conjugate> has principal <complex argument> $-\theta$. Consequently
$$
z^i+\overline z^{\,i}=(e^{-\theta}+e^\theta)e^{i\log r}=2\cosh\theta\,e^{i\log r}.
$$
Taking the <complex modulus> gives $\cosh\theta=1$, hence $\theta=0$. The remaining equation is $e^{i\log r}=i$, so
$$
\boxed{z=e^{\pi/2+2\pi k},\qquad k\in\mathbb Z.}
$$
On the negative real axis, both $z$ and $\overline z$ have principal <complex argument> $\pi$, so the left-hand side has modulus $2e^{-\pi}$, not two. There are no additional solutions there. The printed second term is the principal power of $\overline z$, not the <complex conjugate> of $z^i$.
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