Is my following calculation true?
$e^{a+ib}e^{\overline{a+ib}}=e^{a+ib}e^{a-ib}=e^{2a}$? for a,b real numbers or in general, what is $\overline{{z}^{w}}$ if $z,w$ are complex numbers?
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yes $\color {green}{e^{a+ib}\bar{e^{a+ib}}=e^{a+ib}e^{a-ib}=e^{2a}} $ is true Hint:$$z^w=e^{w \ logz}$$ $$log z=\ln(|z|+\arg(z))$$ Edit :$e^{\bar{z}}=e^{x}\ cos (y)+i\ sin(-y)=e^{x}\ cos (y)-i\ sin(y)$ and $\bar{e^z}={e^{x}\ cos (y)-i\ sin(y)}$
M.H
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Lipschitz: yes notice to definition of exp(z) – M.H Jun 07 '13 at 09:19