# Solving a complex equation including $\Im (z)$

Solve the following complex equation: $$e^{z+5} + e^{\overline z} + i\operatorname{Im}(z) = 6 + i\tfrac{π}{3}$$ I tried using the cartesian form of a complex number $$z$$ meaning $$z=x+iy$$ where $$x$$ and $$y$$ are the real and the imaginary part of $$z$$, respectfully.

Doing some basic calculations (algebra), I conclude that the following system of equations must be valid: \left\{ \begin{aligned} e^x \cos y &= \frac{6}{e^5+1} \\[2pt] e^x \sin y &= \frac{\frac{π}{3}-y}{e^5-1} \end{aligned} \right.

I do not know how to solve this system. I tried to divide the $$2$$ equations but then there is a tany to the LHS and a $$y$$ to the RHS. I also tried to write $$i\operatorname{Im}z$$ as $$e^{\operatorname{Log}(\operatorname{Im}z)}$$ where $$\operatorname{Log}$$ is the principal branch of complex logarithmic function.

• Very mysteriously, I find graphically very many solutions which seem to have the same $x$ coordinate and are periodic in $y$. But that doesn’t make sense… but how can we expect to find the solutions to: $$\tan(y)=\frac{\frac{\pi}{3}-y}{6}\cdot\frac{e^5+1}{e^5-1}$$Which is one of those impossible equations? Where did you get this problem? This seems achievable only numerically Dec 26, 2022 at 11:04
• @FShrike. Almost the same $x$ effectively. One more time for approximation. Cheers :-) Dec 26, 2022 at 12:11
• @FShrike it's from a set of excercises for Complex Analysis,given to a Greek University. I think that the professor had something else in his mind..... Dec 27, 2022 at 22:30

Using @FShrike's comment and cross multiplying we need to find the zeros of function $$f(y)=(\pi -3 y) \cos (y)-k \sin(y) \qquad \qquad k=18\,\frac{ e^5-1}{e^5+1}$$ which, for $$y \in (0,2\pi)$$ shows three roots.
Working the small root, using series $$f(y)=\sum_{n=0}^\infty \frac{\pi \cos \left(\frac{\pi n}{2}\right)-(k+3 n) \sin \left(\frac{\pi n}{2}\right)}{n!}\,y^n$$ Truncating to some order and using power series reversion $$y=\frac{\pi }{k+3}-\frac{\pi ^3}{2 (k+3)^3}+\frac{\pi ^3 \left((k^2+12k+3) \left(9+\pi ^2\right)\right)}{6 (k+3)^5}+\cdots$$
Using only the above terms, the estimate is $$y=\color{red}{0.1503}873$$ while Newton method leads to $$y=0.1503578$$ which is not bad.
Using twice more terms, this would give $$y=\color{red}{0.150357}603$$ while Newton method leads to $$y=0.150357827$$.
We could do the same around $$y=\pi$$ and $$y=2\pi$$ for the same accuracy.
When $$y$$ is known, just compute $$x$$.