Question

Showing \(|e^z|\le 1\text{ if Re}(z)\le 0\), where \(z\) is a complex number.

Answer 1

I tried saying:
\[ |e^z|=\left| \sum_{n=0}^{\infty} \frac{z^n}{n!} \right| \\

=\left| 1+z+\frac{z^2}{2}+\frac{z^3}{3!}+... \right| \le 1+|z|+\frac{|z|^2}{2}+\frac{|z|^3}{3!}+...\] by the triangle inequality hopefully to get somewhere?

Really not sure what to do..

Answer 2

Hm, I was thinking of approaching it this way: Suppose that \(z=x+iy\) for \(x,y\in\Bbb{R}\). Then we have that \[\begin{aligned} |\exp(z)| &= |\exp(x+iy)|\\ &= |\exp(x)\exp(iy)|\end{aligned}\]Now, recall Euler's formula \(\exp(i\theta) = \cos\theta + i\sin\theta\). If we take the modulus of this expression, we see that \(|\exp(i\theta)| = \sqrt{\cos^2\theta+\sin^2\theta}=1\) for any \(\theta\in\Bbb{R}\). Thus, \(|\exp(iy)|=1\). With that, we now see that \(|\exp(z)|=|\exp(x)|\). In terms of \(z\), \(x=\mathrm{Re}(z)\); Thus, \(|\exp(z)|=|\exp(\mathrm{Re}(z))|\). Therefore, if \(|\exp(z)|\leq 1\), it follows that \(|\exp(\mathrm{Re}(z)|\leq 1 \implies -1\leq \exp(\mathrm{Re}(z))\leq 1\). What condition on \(\mathrm{Re}(z)\) makes this possible?

I hope this makes sense! :-)

Answer 3

well it's good to know that since \(\Re\{z\}\in\mathbb{R}\) we know \(e^{\Re\{z\}}>0\) so the lower bound is useless -- we only need to ask \(e^{\Re\{z\}}\le 1=e^0\) which is of course \(\Re\{z\}\le 0\)

Answer 4

Thanks for the response :)!

Answer 5

@oldrin.bataku Yea, I realized I should have mentioned something about that after I posted, but thanks for doing it in my stead. :-)