Ambiguity with complex equation The task is to solve
$$
(z+i)^n + (z-i)^n = 0
$$
with $z \in C$ and for an arbitrary $n$.
My approach is to first rewrite that to
$$
(z+i)^n = - (z-i)^n
$$
and then applying the n-th root to the equation:
$$
\sqrt[n]{(z+i)^n} = \sqrt[n]{- (z-i)^n}
$$
which i believe yields
$$
\sqrt[n]{1} (z+i) = \sqrt[n]{-1} (z-i)
$$
Now I divide both sides by $\sqrt[n]{-1}$ to get
$$
\sqrt[n]{-1} (z+i) = (z-i)
$$
Solving for $z$ I get
$$
z = \frac{\left( \exp\left(\frac{i\pi(1 + 2k)}{n}\right) + 1\right) i}{1 - \exp\left(\frac{i\pi(1 + 2k)}{n}\right)}
$$
with $k \in \{0, 1, \ldots, n-1\}$.
However if I had divided the equation by $\sqrt[n]{1}$ instead of $\sqrt[n]{-1}$ I'd get
$$
z = \frac{\left( \exp\left(\frac{i\pi(1 + 2k)}{n}\right) + 1\right) i}{- 1 + \exp\left(\frac{i\pi(1 + 2k)}{n}\right)}
$$
which is a different result (I think). Thus I must have made a mistake somewhere. However I can't seem to find the problem.
 A: There is no problem at all. Remark that if $z$ is solution $-z$ is also a solution. In facts:
Let $z$ a complex satisfying $(z+i)^n+(z-i)^n=0$ then $(-1)^n(-z-i)^n+(-1)^n(-z+i)^n=0$, simplifying by$(-1)^n$ shows that $-z$ satisfies the initial relation.
————
In addition you may note that:
$$\dfrac{1+e^{ia}}{1-e^{ia}}=\dfrac{2\cos{\frac{a}2}e^{i\frac{a}{2}}}{-2i\sin{\frac{a}2}e^{i\frac{a}{2}}}=i\cot\left(\dfrac{a}2\right)$$
A: Let $z=x+iy$ (where $x,y\in\Bbb R$). The equation becomes
$$[x+(y+1)i]^n+[x+(y-1)i]^n=0$$
This implies that
$$|x+(y+1)i|=|x+(y-1)i|$$
that is,
$$(y+1)^2=(y-1)^2$$
So $y=0$. Now, we know that $z=x$ is real.
If $x=0$ the equation holds for odd $n$; in other words, if $n$ is odd, $z=0$ is a solution.
Otherwise
$$(x\pm i)^n=(x^2+1)^{n/2}e^{\pm in\arctan(1/x)}$$
Then the nonzero solutions are given by the equation
$$2n\arctan(1/x)\equiv \pi\pmod{2\pi}$$
or
$$z=x=\cot\left(\frac{\pi+2k\pi}{2n}\right), k=0\ldots n-1$$
A: I don't think you should have divided both sides by anything after the equation $$
\sqrt[n]{1} (z+i) = \sqrt[n]{-1} (z-i).
$$
 If we write $w_n=\sqrt[n]{1}$ and $w'_n=\sqrt[n]{-1},$ and solve for $z$ in the equation, then we get $$z=\frac{i(w_n+w'_n)}{w'_n-w_n}.$$ You may now simplify as before.
