Are there any solution for a,b,c,d such that $(a+bi)^{n}+(c+di)^{n}=2i$ Are there any solution for a,b,c,d such that $(a+bi)^{n}+(c+di)^{n}=2i$.
With a,b,c,d,n are positive integer numbers and $a+bi, c+di$ are complex numbers
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I just have started learning about comlex number, so I do not know too much about it
And if you can find all solution, please show me how to find it.
 A: For $\color{blue}{n=1}$:
$$\begin{align}
(a+bi)^{1}+(c+di)^{1}=(a+c)+(b+d)i=2i \\
\implies a+c=0\quad &(\text{equating real parts})
\end{align}$$
has no solutions because $a,c\ge1$. There would be solutions if $a,b,c,d$ were non-zero positive or negative integers.
For $\color{blue}{n=2}$:
$$\begin{align}
(a+bi)^{2}+(c+di)^{2}=(a^2-b^2+c^2-d^2)+(2ab+2cd)i=2i \\
\implies 2ab+2cd=2\quad &(\text{equating imaginary parts})
\end{align}$$
has no solutions because $a,b,c,d\ge1\implies2ab+2cd\ge4$. 
For $\color{blue}{n\ge3}$:
Firstly, note that $|(x+yi)^n|=|x=yi|^n$ for any complex number $x+yi$.
Then $(a+bi)^n+(c+di)^n+(-2i)=0$ so thinking of the numbers $(a+bi)^n$, $(c+di)^n$ and $-2i$ as vectors in the complex plane, they must form a triangle that closes. Hence, the side lengths must obey the triangle inequality.
In particular, if we assume WLOG that $|(a+bi)^n|\ge|(c+di)^n|$ then we must have
$$|(a+bi)^n|-|(c+di)^n|\le|2i| \implies |a+bi|^n-|c+di|^n\le2 \tag{1}$$
Assuming that it is possible to find a solution to (1) with $|a+bi|>|c+di|\text{ for }n\ge3$, and noting that $a,b,c,d$ are all positive integers:
$$\begin{align}
|a+bi|^n-|c+di|^n &= (\sqrt{x})^n-(\sqrt{y})^n\text{ for some }x>y\ge2 \\
                  &= x^{3/2}-y^{3/2} \\
                  &\ge 3^{3/2}-2^{3/2} > 5.1 - 2.9 > 2 
\end{align}$$
So it is not possible to find a solution to (1) where $|a+bi|\ne|c+di|$.
We can conclude that $|a+bi|=|c+di|$ and write $a+bi,c+di$ in polar form as:
$$a+bi=r(\cos\alpha+i\sin\alpha),c+di=r(\cos\beta+i\sin\beta)$$
[added] where $r=|a+bi|=\sqrt{a^2+b^2}\ge\sqrt2$ because $a,b\ge1$.
Then by de Moivre's formula,
$$(a+bi)^{n}+(c+di)^{n}=r^n(\cos{n\alpha}+i\sin{n\alpha})+r^n(\cos{n\beta}+i\sin{n\beta})=2i $$
so by equating real parts
$$\cos{n\alpha}+\cos{n\beta}=0 \tag{2}$$
and by equating imaginary parts
$$r^n\sin{n\alpha}+r^n\sin{n\beta}=2 \tag{3}$$
Condition (2) requires that $$n\alpha+n\beta=\frac{\pi}{2}+k\pi \tag{4}$$ (rule out $n\alpha-n\beta=\pi+2k\pi$), so the points are symmetrically placed on either side of the imaginary axis, and condition (3) then reduces to
$$r^n\sin{n\alpha}=r^n\sin{n\beta}=1 \tag{5}$$ 
Finally, we must have
$$\sin{n\alpha}=\sin{n\beta}=\frac{1}{\sqrt2} \implies r^n=\sqrt2$$
which has no solutions, because $r\ge\sqrt2$ and $n\ge3$.
So, there are n solutions for $n\ge3$. There would be no solutions even if $a,b,c,d$ were allowed to be positive or negative integers.
