0
$\begingroup$

I'm reading Conway's complex analysis book and on page 116 he writes:

Why is the last equality true?

$\endgroup$
1
  • 4
    $\begingroup$ $|e^{x+iy}|=e^x$. $\endgroup$ May 24, 2016 at 18:39

2 Answers 2

4
$\begingroup$

$ |e^{iRe^{i\theta}}| = |e^{{iR(\cos\theta + i\sin\theta)}}| = |e^{{iR\cos\theta}}e^{-R\sin\theta}| = |e^{-R\sin\theta}|, $

since $R\cos\theta$ is real, hence $|e^{{iR\cos\theta}}| =1$

$\endgroup$
0
$\begingroup$

Because for $\;x,y\in\Bbb R\;$ :

$$e^{x+iy}=e^xe^{iy}\implies\left|e^{x+iy}\right|=|e^x|\,|e^{iy}|=e^x$$

since $\;|e^{iy}|=|\cos x+i\sin x|=\sqrt{\cos^2x+\sin^2x}=1\;$

And in your case:

$$e^{iRe^{i\theta}}=e^{iR(\cos\theta+i\sin\theta)}=e^{-R\sin\theta}e^{iR\cos\theta}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.