Is it true that $|e^z|\le e^{|z|}$ for all $z \in \mathbb C$?

Is it true that $|e^z|\le e^{|z|}$ for all $z \in \mathbb C$?

I believe so as $x \le \sqrt{x^2+y^2}$ but the way the question is worded suggests it is not the entire complex plane.

Solution 1:

$$\left|e^z\right|=\left|e^{x+iy}\right|=\left|e^x\right|\left|e^{iy}\right|=e^x\leq e^{|z|}$$ since $x\leq|z|$ and the exponential function is monotonic increasing.

Solution 2:

It is correct. For all $z \in \Bbb C$, $\text{Re} \, z \le |z|$ and therefore $$ |e^z| = e^{\text{Re} \, z} \le e^{|z|} $$

Solution 3:

Consider the Taylor expansion, and apply the regular absolute value inequality to it for each finite term, then take the limit.

Solution 4:

Recall that we have $$ e^z = e^{\Re z}e^{i\Im z} $$ As $|e^{i\Im z}| = 1$, we have $|e^z| = e^{\Re z}$. As $\Re z \le |z|$ and the exponential is monotone on $\mathbf R$, we have $$ |e^z| = e^{\Re z} \le e^{|z|}. $$

Solution 5:

$|e^{a+bi}|=|e^a|\cdot|e^{bi}|=|e^a|=e^a$. On the other hand, $e^{|a+bi|}=e^{\sqrt{a^2+b^2}}$, and as you noted, $a\leq\sqrt{a^2+b^2}$ for every $a\in\mathbb{R}$. Therefore, $|e^z|\leq e^{|z|}$ for every $z\in\mathbb{C}$, as $e^x$ is monotonically increasing for $x\in\mathbb{R}$ and $a,\sqrt{a^2+b^2}\in\mathbb{R}$ with $a\leq\sqrt{a^2+b^2}$.