Complex conjugates?

Question

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.

Answer 1

Hint: $z$ and $w$ are complex conjugates if and only if $zw$ is a real number and $|z|=|w|$. You might wish to check both conditions for $z=\exp(1/u)$ and $w=\exp(1/\bar u)$ and any nonzero complex number $u$.

Edit To compute $|z|$ and $|w|$, you could start from the fact that, for every real numbers $a$ and $b$, $\exp(a+\mathrm ib)=\exp(a)\exp(\mathrm ib)$. Since $\exp(\mathrm ib)=\cos(b)+\mathrm i\sin(b)$ has absolute value $1$, $\exp(a+\mathrm ib)$ has absolute value $\exp(a)$. A second step could be to write $z=\exp(1/u)$ as $z=\exp(a+\mathrm ib)$, hence to write $1/u$ as $a+\mathrm ib$ for $u=1-\mathrm ix$, then to do the same for $1/\bar u=c+\mathrm id$, and finally to check that $a=c$ (hence $|z|=|w|$) and $b+d=0$ (hence $zw$ is a real number).