$\gamma$ is the straight line segment joining $0$ to $\pi - i \pi$. Evaluate: $\int_{\gamma} e^{\bar{z}} dz$.

Asked Nov 08 '22 at 11:30

Active Mar 14 '23 at 06:08

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$\gamma$ is the straight line segment joining $0$ to $\pi - i \pi$. I need to evaluate: $\int_{\gamma} e^{\bar{z}} dz$.

Is that correct?

$$\int_{\gamma} e^{\bar{z}} dz = \int_{0}^{1} \exp \left[ (\pi + i \pi)t \right] (\pi - i \pi)dt =$$ $$ =(\pi - i \pi) \int_{0}^{1} \exp \left[ (\pi + i \pi)t \right]dt = (\pi - i \pi) \int_{0}^{1} e^{(\pi + i \pi) t}dt = $$ $$ =(\pi - i \pi) \left[\frac{1}{\pi + i \pi} \cdot e^{(\pi + i \pi) t} \right]^{1}_{0} = (...) = (e^{\pi}+1)i$$

edited Mar 14 '23 at 06:08

FD_bfa

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asked Nov 08 '22 at 11:30

thefool

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$dz=(\pi-i\pi)dt$ not $(\pi+i\pi)dt$. – David Nov 08 '22 at 12:29
Oh, ok. But only $dz$, right? Because in the $\exp$ function we got conjugate. – thefool Nov 08 '22 at 12:49
@geetha290krm excellent pun... – Jean Marie Nov 08 '22 at 12:50
Does this answer your question? Evaluate $\int \bar z dz$ – Jean Marie Nov 08 '22 at 12:51
It is outside exponent – thefool Nov 08 '22 at 12:54
$e^{\pi t}(\cos \pi t + i\sin\pi t)$ in the last line.... you dropped the $i.$ You can proceed with integration by parts. However, you don't have to work that hard... $(\pi -i\pi)\int_0^1 e^{(\pi+i\pi)t}\ dt = (\pi - i\pi) \frac {e^{(\pi+i\pi)t}}{\pi + i\pi}|_0^1$ – user317176 Nov 08 '22 at 18:27
It's done. I will update question. Can you check if it's ok? @DougM – thefool Nov 08 '22 at 19:05
But there can't be $t$ according to your last comment. – thefool Nov 08 '22 at 19:37
$ = (\pi - i \pi)\left[\frac{e^{\pi + i \pi}}{\pi + i \pi} - \frac{1}{\pi + i \pi} \right] =(1- i)\left[\frac{e^{\pi + i \pi}}{1 + i} - \frac{1}{1 + i} \right] = \frac{-2i(-e^{\pi}-1)}{2} = (e^{\pi}+1)i$ @DougM – thefool Nov 08 '22 at 19:43
Sorry, no t... $\pi(1-i)(\frac {e^{\pi(1+i)t}}{\pi(1+i)})|_0^1 = (1-i)(\frac {e^{\pi(1+i)}-1}{1+i}) = (1-i)^2(\frac {e^{\pi(1+i)}-1}{(1+i)(1-i)}) =(-2i)(\frac {e^{\pi(1+i)}-1}{2}) = -i(e^{\pi(1+i)}-1) = i(1-e^{\pi(1+i)})$ – user317176 Nov 09 '22 at 01:17
So it's the same since $e^{\pi i} = -1$. Is that correct? @DougM – thefool Nov 09 '22 at 01:20

$\gamma$ is the straight line segment joining $0$ to $\pi - i \pi$. Evaluate: $\int_{\gamma} e^{\bar{z}} dz$.

0 Answers0