<s>[INST] Complex Numbers Identity Let $a,b$ be non-zero complex numbers with:
$$a2^{|a|}+b2^{|b|} = (a+b)2^{|a+b|}$$
Show that $a^6 = b^6$.
I gave it a go, but to no avail.
 [/INST] First note that $a+b=0$ always gives a solution for which $a^6=b^6$.
If $a$ and $b$ are real, we can assume without loss of generality that $a=1$. We get $2+b2^{|b|}=(1+b)2^{1+|b|}$, so $2^{1-|b|}+b=2(1+b)$, so $2^{1-|b|}-b=2$. By piecewise analysis, we find $b=0$ or $b=-1$. But $a$ and $b$ are required to be non-zero, so we have $a+b=0$.
From now on, assume $a+b\neq0$. By rotation, we may assume without loss of generality that $a+b$ is positive and real. Then $a2^{|a|}+b2^{|b|}=(a+b)2^{|a+b|}$ is also positive and real. Since $a$ and $b$ are not both real, neither is real. It follows that $2^{|a|}=2^{|b|}$, so $|a|=|b|$.
Write $a=re^{i\theta}$ and $b=re^{-i\theta}$. We get $a+b=2r\cos(\theta)>0$. So the equality becomes $$2r2^r\cos(\theta)=2r\cos(\theta)2^{2r\cos(\theta)},$$ so $r=2r\cos(\theta)$. This gives $\theta=\frac{\pi}3$, so $a^6=b^6$.
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