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Show that $|e^z| \le 1$ if $Re [z] \le 0$

I know that $z=a+bi$ so $|e^{a+bi}|$ and $a$ represents the radius and $b$ represents the angle $\theta$ but I think I need to convert $e^z$ to complex form in order to take the modulus? What I don't really understand is what $e^z$ means?

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It's a useful fact that $$|e^{z}| = |e^{a+bi}| = |e^{a}e^{bi}| = |e^{a}||e^{bi}| = |e^{a}| = e^{a}$$

Since $a = \Re z$, if $a \leq 0$ then $e^{a} \leq 1$, which in turn means $|e^{z}| \leq 1$.

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    I understand everything you wrote, but can the imaginary part, $|e^{bi}|$ be large enough making $|e^z|$ greater than $1$? – 2017-02-05
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    @idknuttin No, the point is that for real $b$, we always have $|e^{bi}|=1$ by Euler's formula. – 2017-02-05
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    To follow up on @angryavian's point, $|e^{bi}| = |\cos b + i\sin b| = \sqrt{\cos^{2}b + \sin^{2}b} = \sqrt{1} = 1$ – 2017-02-05
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Set $z=x+\mathrm iy$. Then we have $|\mathrm e^z|=\mathrm e^x|\mathrm e^{\mathrm iy}|=\mathrm e^x=\mathrm e^{\operatorname{Re} z}$ because $\mathrm e^x>0$ for every real $x$. Since $x=\operatorname{Re} z \leq 0$, we have $|\mathrm e^z|=\mathrm e^x \leq 1$ where we used that the exponential function is monotone.