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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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1 Answer 1

up vote 2 down vote accepted

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 $\bar{e^z}={e^{x}\ cos (y)-i\ sin(y)}$

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Thank you! then i got it. –  Xin Wang Jun 7 '13 at 9:15
Lipschitz: yes notice to definition of exp(z) –  Maisam Hedyelloo Jun 7 '13 at 9:19

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