2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.