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.

  • $\begingroup$ The parametrization of a line joining $z_1,z_2$ in $\Bbb C$ is given by $\varphi(\lambda)=\lambda z_1+(1-\lambda)z_2$ where $\lambda\in [0,1]$ $\endgroup$ – Prasun Biswas Apr 20 '18 at 13:51
  • $\begingroup$ The line starts at $z_1=4$ to according to your formula when $\lambda=0$ you get $z_2$. But anyway $\gamma_{(t)}=4\cdot (1-t) +(4-3i)t=4-3i\cdot t$ $\endgroup$ – bp7070 Apr 20 '18 at 13:55
  • $\begingroup$ The line joining $z_1,z_2$ is the same as the line joining $z_2,z_1$ (just the orientation is reversed, so the sign of the integral is flipped). Since you're working on the absolute value of the integral, it shouldn't matter. $\endgroup$ – Prasun Biswas Apr 20 '18 at 13:58
  • $\begingroup$ OK, but does it help here? $\endgroup$ – bp7070 Apr 20 '18 at 14:01

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.

  • $\begingroup$ Thanks, where does the 3 before the integral come from? $\endgroup$ – bp7070 Apr 20 '18 at 15:01
  • $\begingroup$ It comes from nowhere. It's a mistake, I'll edit, thanks. $\endgroup$ – zhw. Apr 20 '18 at 15:09

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