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Is $e^{1\over 1-ix}$ the complex conjugate of $e^{1\over 1+ix}$? Is there a simple rule to compute complex conjugates without having to find $a+ib$ form? Thanks.

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    "Is there a simple rule to compute complex conjugates?" - yes, replace all instances of $i$ with $-i$. For instance, the conjugate of $\exp(a-ib)$ is $\exp(a-(-i)b)=\exp(a+ib)$...2011-10-16
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    @J.M.: I think that ought to be an answer.2011-10-16
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    @J.M.isn'tamathematician Does your comment hold in any case, no matter whether your complex number is in the form $a+bi$, $e^{i\delta}$ or other? I calculated that the complex conjugate of $(a+bi)e^{i\delta}$ is indeed $(a-bi)e^{-i\delta}$.2017-06-26
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    @Karlo, it should. Proving it is so is a nice exercise.2017-07-26

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