Limit with complex numbers

Question

I've never calculated limits with complex numbers before. Why does

$$ \lim_{z\rightarrow \infty} \frac{e^{3iz}-3e^{iz}}{z^3} =0?$$

This is contrary to my intuition, since exponentials grow faster than powers.

Answer 1

Note that for $t$ real we have

$$ e^{3it}-3e^{it} =\cos(3t)-3\cos(t)+i(\sin(3t)-3\sin(t))$$

which is a bounded function:

$$\left| e^{3it}-3e^{it} \right| \leq \left| \cos(3t) \right| +\left|-3 \cos(t) \right| + \left| i\sin (3t) \right| +\left| -3i \sin(t) \right| \\ \leq 1+3+1+3=8$$

The intuition that exponentials grow faster than polynomials is true for real exponentials. The complex exponentials are bounded in the imaginary direction.

Answer 2

Suppose $t\to+\infty$ on the real axis and $z=-it$. Then we have $$ \frac{e^{3iz}-3e^{iz}}{z^3} = \frac{e^{3t} - 3e^t}{it^3} $$ and that does not approach $0$.

This complex-valued function has an essential singularity at $\infty$.