Show that $$\left | \int_{C}^{ } \frac{e^z}{z^2+2i}dz \right |\leq 2\pi e^2$$ when $C$ is the line connecting between $z_1=4$ to $z_2=4-3i$.
I tried to find parametrization $\gamma_{(t)}=4-3it, t\in[0,1]$ and use it as
$$\left | \int_{0}^{1} \frac{e^{(4-3it)}}{{(4-3it)}^2+2i}\cdot -3i \cdot dz \right |\leq \int_{0}^{1} \frac{|e^{(4-3it)}|}{{|(4-3it)}^2+2i|}\cdot |-3i| \cdot|dz|$$
but I can't see how it helps.
I assume the answer is related somehow to a circle because of the $\pi$ but I can't see the connection.
Any hint or answer will be appreciated.
I parameterized $C$ as $4-it, 0 \le t \le 3.$ We then have the expression bounded above by
$$e^4 \int_0^3\left | \frac{1}{16-t^2 +i(2-8t)}\right|\, dt \le e^4 \int_0^3 \frac{1}{16-t^2 }\, dt \le e^4 \cdot 3\cdot \frac{1}{7}.$$
So is the last number $\le 2\pi e^2?$ Indeed it is. Cancel the $e^2$ and you're left wondering if $e^2\cdot 3/7 \le 2\pi.$ That's true by a mile, and you don't need a calculator.