Is the Mapping (x,y) -> (u^2-v^2,2uv) Conformal Everywhere Except the Origin?

[Ackbach]

Here is this week's POTW:

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I am indebted to Philip Exeter's Math Problems for the following problem. Prove that the mapping $(x,y)\mapsto (u^2-v^2,2uv)$ is conformal everywhere except the origin.

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[Ackbach]

No one answered this week's POTW. Here is my solution:

Spoiler

Note that if we let $z=u+iv$, then $z^2=u^2-v^2+2iuv$, and is an analytic function. Its derivative is $2z$, and so we see that $z^2$ is conformal everywhere except at the origin. But the real part of $z^2$ is the $x$-component of the function given, and the imaginary part of $z^2$ is the $y$-component of the function given. Therefore, the function given is conformal everywhere except at the origin.