Integral calculations using Cauchy's Integral formula

  • Thread starter Susanne217
  • Start date
  • #1
317
0

Homework Statement



Calculate the integral

Given [tex]\int_{C} \frac{e^z}{\pi i - 2z} dz = \int_{C} \frac{e^z}{z-\frac{\pi i}{2}} dz} [/tex]

using Cauchy integral formula.

Homework Equations



What I know

[tex]\frac{1}{2\pi i} \int_{C} \frac{f(z)}{z-\zeta} = 2\pi i f(\zeta)[/tex]

The Attempt at a Solution



This in my little girly mind amounts to

[tex]2\pi i f(\frac{\pi i}{2}) = \int_{C} \frac{e^z}{z-\frac{\pi}{2}i} dz \Rightarrow \int_{C} \frac{e^z}{z-\frac{\pi}{2}i} dz = 2 \pi \cdot (i) \cdot (i) = -2\pi[/tex]

But people who are wiser than me says to me "Susanne your result is wrong!". Could someone please point out my mistake?

thanks Susanne
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
e^z/(i*pi-2z) isn't equal to e^z/(z-i*pi/2). It's equal to (-1/2)*e^z/(z-i*pi/2). You can't just drop the (-1/2) constant.
 
  • #3
317
0
e^z/(i*pi-2z) isn't equal to e^z/(z-i*pi/2). It's equal to (-1/2)*e^z/(z-i*pi/2). You can't just drop the (-1/2) constant.

Dick,

You are a genius my man. I simply couldn't see that :D
 

Related Threads on Integral calculations using Cauchy's Integral formula

Replies
4
Views
1K
  • Last Post
Replies
3
Views
4K
Replies
1
Views
927
  • Last Post
Replies
2
Views
12K
  • Last Post
Replies
2
Views
890
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
1K
Top