1. ## complex analysis questionnn

i have put up the question as a jpeg attachment since i dont know how to type all the math symbols.
i would appreciate any help i am quite stuck it is one of the review questions for my monday exam.

2. Is...

$\displaystyle \int_{\gamma} f(z)\ dz = 2 \pi i \sum_{k} r_{k}$

... where the $r_{k}$ are the residues of the poles inside $\gamma$. In this case the poles are $z=1$ and $z=2$, both inside $\gamma$...

Kind regards

$\chi$ $\sigma$

3. Yes.

4. what i was wondering was whether I could simply expand z = x+iy throughout the integral and then solve it that way by splitting the real and imaginary parts or whether there was a shortcut/better way to go about doing it?

Thanks a lot!

5. Originally Posted by shaheen7
what i was wondering was whether I could simply expand z = x+iy throughout the integral and then solve it that way by splitting the real and imaginary parts or whether there was a shortcut/better way to go about doing it?

Thanks a lot!
As Chisigma mentioned in post number 2 Cauchy Residue theorem is the "shortcut way" of doing this problem. Have you learned this theorem?

6. yes i have studied the theorem i am just a bit confused about how to go about getting the residue for this particular problem.

7. Originally Posted by shaheen7
yes i have studied the theorem i am just a bit confused about how to go about getting the residue for this particular problem.
Since both of the poles are simple the resides can be calculated as follows

$\displaystyle f(z)=\frac{e^z}{(z-1)(z-2)}$

The residue at $z=1$

$\displaystyle \text{Res}(f,z=1)=\lim_{z\to1}(z-1)\frac{e^z}{(z-1)(z-2)}=\lim_{z \to 1}\frac{e^z}{(z-2)}=-e$

In general an nth order pole's residue can be calculated using the formula from post #4 in this link

http://www.mathhelpforum.com/math-he...tml#post544571

8. so the answer would simply be 2(pi)i multiplied by (-e)(e2) which are the residues? is that it?

9. Originally Posted by shaheen7
so the answer would simply be 2(pi)i multiplied by (-e)(e2) which are the residues? is that it?
it would be the sum of the residues

$2\pi i ( -e+e^2)$