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.

  • 1
    $\begingroup$ 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 $\endgroup$
    – FShrike
    Dec 26, 2022 at 11:04
  • $\begingroup$ @FShrike. Almost the same $x$ effectively. One more time for approximation. Cheers :-) $\endgroup$ Dec 26, 2022 at 12:11
  • $\begingroup$ @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..... $\endgroup$ Dec 27, 2022 at 22:30

1 Answer 1


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 $x$.


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