Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question : Given $\text{Re}(z) \le 0$ prove that $|e^z| \le 1$.

Try: $z=x+yi$, it's given that $x \le 0$.

$$|e^{z}| = |e^{x+yi}|=|e^xe^{yi}|=e^x|e^{yi}|,$$ with $e^x \le e^0$ because $f(x)=e^x $ is a increasing function everywhere.

What's next? What can I say about $|e^{yi}$| ?

share|cite|improve this question
up vote 4 down vote accepted

$$\vert e^{iy} \vert = \underbrace{\vert \cos(y) + i \sin(y) \vert = \sqrt{\cos^2(y) + \sin^2(y)}}_{\because y \in \mathbb{R}} = 1$$

share|cite|improve this answer
Im really sad I need this site for such a silly answer ;) – Applied mathematician Nov 18 '12 at 21:03
@Hempo It is fine :). It happens to everyone :-). Also, it is worth noting that $$\left \vert \cos(y) + i \sin(y) \right \vert = \sqrt{\cos^2(y) + \sin^2(y)}$$ is true only when $y \in \mathbb{R}$ and not in general when $y \in \mathbb{C}$. – user17762 Nov 18 '12 at 21:10
Thanks for your help Marvis! – Applied mathematician Nov 18 '12 at 22:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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