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Let $f:(a+bi) \to (a^2+b)+ai$

Is there an expression for f such that does not use functions $\Re$ and $\Im$?

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closed as off-topic by Watson, Rory Daulton, Jonas, Hirshy, 91500 Mar 26 at 11:26

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You can try substituting in $a = \frac{1}{2}(z + \bar z)$ and $b = \frac{1}{2i}(z - \bar z)$ and seeing if your expression simplifies. – Rahul Dec 21 '10 at 17:23
My bet would be no, but I wouldn't know how to prove it. – Noldorin Dec 21 '10 at 17:25
Following Rahul's suggestion, $f(z) = \frac14 z^2 + \frac12 |z|^2 + \frac14 \overline{z}^2 + i\overline{z}$. – Dave Radcliffe Dec 21 '10 at 18:09
Seems like using $\overline{z}$ is cheating if we are not allowed Im. – Ross Millikan Dec 21 '10 at 18:13
What is an "expression"? If you allow $\overline{z}$, then you can express Re and Im in terms of it as Rahul writes. If you allow $|z|$ then $|z|^2/z=\overline{z}$ for all nonzero $z$. If you limit yourself to analytic functions, then the answer is "no" as lhf explains. – David Speyer Dec 21 '10 at 20:56

No expression as an analytic function because it does not satisfy the Cauchy-Riemann equations.

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