So, I came across this equation and started playing around with it to see if I could solve it exactly. The first thing I noticed is that if I assume $z$ lies in the real or the imaginary axis the only possible solution is $0$. But this cannot be a solution because the right hand side of the equation is undefined at this point. So we can assume without loss of generality that both the real and imaginary parts of $z$ are non-zero.

The next thing I noticed is a symmetry in the equation. If $z$ is a solution, then so is $-z$. We can see this by writing $1=[(-1)^{n}(-1)][(-1)^{n}(-1)^{-1}]$, substituting in the left hand side of the equation, moving $(-1)^{n}(-1)^{-1}$ to the right hand side and rearranging. I actually "reversed engineered" this algebraic manipulation by substituting $-z$ in the equation and playing with it for a while. So far so good.

Then I wanted to try and see if I could find any actual solutions. After a while, I realized that I could try to solve an equivalent system of equations. Two non-zero complex numbers are equal if and only if their absolute values are equal and their arguments are congruent modulo $2\pi$. So I could write the original equation as:

\begin{cases} \vert z\vert^{2n}\vert Re(z)\vert=\vert Im(z)\vert \\ \arg(z^{2n}Re(z))\equiv_{2\pi}\arg(Im(z)) \end{cases}

The first equation is not very useful at the moment so let's just ignore it. The second one, however, simplifies to $2n\arg(z)\equiv_{2\pi} \arg(Im(z))$ by making use of the fact that $\arg(zw)\equiv_{2\pi}\arg(z)+\arg(w)$. There are two cases to consider depending of the sign of $Im(z)$. Let's assume it's positive so that the right hand side of the congruence becomes $0$.

To make things easier to work with, we rewrite the expression in terms of equality as $2n\arg(z)=2k\pi$. Rearranging yields $\arg(z)=\frac{k\pi}{n}$. Now I know for a fact that there's a problem with these answers. Take for instance $n=2$. According to my calculations, any solution $z$ to the original problem with this value of $n$ should solve $\arg(z)=\frac{k\pi}{2}$ (A). I know that this isn't necessarily true in the other direction (unless the other equation in the system is satisfied), but I only need the first implication anyway.

(A) says that the solutions should lie on either the real or the imaginary axis, which we know can't happen. So I thought "okay, there are no solutions for $n=2$ and $Im(z)>0$ because all the candidates fail to satisfy the original problem". But here's the problem. I checked with Wolphram Alpha and there should be two distinct solutions: $\pm\frac{1-i}{\sqrt{2}}$. Since one of these has positive imaginary part it should have been among my solution candidates.

The question is: where did I go wrong? I've went over the calculations and as far as I can tell there's nothing wrong with what I did. I can only think that maybe I messed up when I threw the argument into the mix, but even then I don't see any problem with the reasoning. I mean, there's obviously something wrong in the math, but I can't figure out what it is.

It's probably something really obvious so I apologize in advance for my idiocy, but I really would appreciate your help in figuring out where the mistake is.

Edit: If I consider the case when the imaginary part is negative I get the arguments of the two solutions that Wolphram Alpha gives. But what I find strange is that, technically, the solution with positive imaginary part should not be considered valid. What am I missing here?

Another edit: I think I now know where I messed up. When I simplified the congruence equation I completely forgot to take into account the sign of the real part. In total, I should have four cases to consider. One per each combination of signs. I'll come back to it tomorrow to make sure since it's getting late, but I'm pretty sure that is exactly the problem.


Take at once the polar form $z=re^{i \theta}$:

$$r^ne^{in \theta} \ r \cos \theta=\dfrac{1}{r^n} \ e^{-i n \theta}r \sin \theta$$

$$r^{n+1}\cos \theta \ e^{in \theta}=\dfrac{1}{r^{n-1}}\sin \theta \ e^{-i n \theta}\tag{1}$$

Then identify the moduli and the arguments (mod $2 \pi$).

For the moduli, beware of the signs ($\cos \theta$ or $\sin \theta$ can be negative). As the cases $\theta=k \pi/2$ are excluded (thy would leas in (1) to a side equal to $0$ and the other side $\ne 0$), you have to consider 4 cases

  • a) $ \ \ 0<\theta<\pi/2$: the upsaid identification gives $ \ \ r^{2n}=\tan \theta$, $ \ \ 2n \theta=k 2 \pi. $ We can therefore conclude that there are $n-1$ solutions in this case with :

$$\theta_k=\dfrac{k 2 \pi}{n} \ (k=1,2, \cdots k_0), \ \text{and} \ r=(\tan \theta_k)^{\frac{1}{2n}}$$

where $k_0$ is the greatest integer $k$ such that $\dfrac{k}{n}<\dfrac{1}{2}$.

  • b) $ \ \ \pi/2<\theta<\pi$: write (1) under the form:

$$r^{n+1}(-\cos \theta)e^{i \pi} \ e^{in \theta}=\dfrac{1}{r^{n-1}}\sin \theta \ e^{-i n \theta}\tag{2}$$

and proceed as in case a).

  • c) $ \ \ \pi<\theta<3\pi/2$ : same as case a)

  • d) $ \ \ 3\pi/2<\theta<4\pi$ : same as case b).

  • $\begingroup$ Seconds before I recieved the notification for your comment I realized that I had messed up the signs . In my calculations I had implicitly assumed the real part to be positive. I'm pretty sure that's the reason why I wasn't getting the right values. Thank you for your answer though. That's probably a better way to anaylize the problem. $\endgroup$ – Modesto Rosado Sep 19 '20 at 6:56
  • $\begingroup$ OK, I had added some more stuff to my answer. $\endgroup$ – Jean Marie Sep 19 '20 at 6:59

We have that for $z\neq 0 \implies \Re(z)\neq 0$

$$z^{n}\Re(z)=z^{-n}\Im(z) \iff z^{2n}=\frac{\Im (z)}{\Re (z)}\in \mathbb R$$

$$\rho^{2n}\left(\cos(2n\theta)+i\sin(2n\theta)\right)=\frac{\sin \theta}{\cos \theta}$$

therefore we have $\sin(2n\theta)=0 \iff 2n\theta=k\pi$ and then


that is for $k=2h$ even


for $k=2h+1$ odd



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