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There exist a set of $x,y\in\mathbb{C}$ that $|x^2|+|2xy|+|y^2|=|(x-y)^2|$. Find the imaginary part of $\frac{x}{y}$.

So I tried to use the triangular inequality and $|x||y|$=$|xy|$ but I cant get anywhere from here. Can anyone provide a hint to do this? Thank you!

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    Hint: dividing both sides by $y^2$ gives a relation in $z=x/y$. – 2017-01-31
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    @dxiv Or rather, one should divide by $|y^2|$? – 2017-01-31
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    @SimplyBeautifulArt Right, that's what I meant of course ;-) Thanks. – 2017-01-31
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    Yeah I tried this way but cant get to anywhere, i get $|x/y|^2 + 2 |x/y| +1= | (x/y)^2 -2x/y +1| .$ Is there something behind this? – 2017-01-31
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    RHS is better written as $|x/y-1|^2=(z-1)(\bar z -1)$. – 2017-01-31

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We may assume $y \ne 0$, because otherwise the number $\dfrac{x}{y}$ will be undefined.

If $x=0$, then $\dfrac{x}{y}=0$ and its imaginary part is also $0$.

Let's now assume that both $x$ and $y$ are non-zero. The left hand side is $$ |x^2|+|2xy|+|y^2|=|x|^2+2|x||y|+|y|^2 $$ while the right hand side is $$ |(x-y)^2|=|x-y|^2=(x-y)(\bar{x}-\bar{y})=|x|^2-x\bar{y}-\bar{x}y+|y|^2, $$ therefore if we denote by $\alpha$ and $\beta$ the arguments of $x$ and $y$, respectively, the original equation becomes $$ 2|x||y|=-x\bar{y}-\bar{x}y=-|x||y|e^{i(\alpha-\beta)}-|x||y|e^{-i(\alpha-\beta)}=-2|x||y|\cos(\alpha-\beta) $$ The equation is then equivalent to $$ 2|x||y|[1+\cos(\alpha-\beta)]=0, $$ i.e. $\cos(\alpha-\beta)=-1$. In particular $\sin(\alpha-\beta)=0$. It follows that $$ \dfrac{x}{y}=\dfrac{|x|}{|y|}e^{i(\alpha-\beta)}=-\dfrac{|x|}{|y|}, $$ and the imaginary part of $\dfrac{x}{y}$ is $0$.