# Math Help - 2 more questions in Residue Theorem-Complex

1. ## 2 more questions in Residue Theorem-Complex

1. How can I calculate the residue in z=1 of the next function:
$f(z)=(2z-1)e^{\frac{z}{z-1}}$ ??
I've tried using Laurent expansion but couldn't figure out how to expand this function into a series (using the known expansion of the exponent don't help us here... ).

2. Calculate the next Integral:
$\frac{1}{2\pi i} \int_{|z|=1} \frac{z^{n-1}}{3z^n-1} dz$.

My problem with this question is that we have n singularities and they are all inside our contour... I've tried to paramterize but without any success..
Hope you'll be able to help me

Thanks in advance!

2. I believe the answer to the second part of your question is 1/3....

$(1/2 \pi i) \oint\frac{1/3z^{n-1}dz}{z^n-1/3} = \sum_{k=0}^{n-1}\frac{1/3(\zeta _{k})^{(n-1)/n}}{n(1/3)^{(n-1)/n}}e^{2\pi ik(n-1)/n}$ with $\zeta _k = (1/3)^{1/n}e^{2\pi ik/n}$ for k=0,1,2.....,n-1

Basically just sum over the residues evaluated at the n roots of 1/3...

Hope this helps...

3. Originally Posted by WannaBe
1. How can I calculate the residue in z=1 of the next function:
$f(z)=(2z-1)e^{\frac{z}{z-1}}$
$(2z-1)e^{z/(z-1)}=(2z-1)e\sum_{n=0}^{\infty}\frac{1}{n!(z-1)^n}$

$=2e\sum_{n=0}^{\infty}\frac{(z-1)^{1-n}}{n!}+e\sum_{n=0}^{\infty}\frac{1}{n!(z-1)^n}$

and the coefficient on the 1/(z-1) term is:

$2e(1/2)+e=2e$

which is consistent with the numerical result:

Code:
In[13]:=
NIntegrate[(2*z - 1)*Exp[z/(z - 1)]*(1/4)*
I*Exp[I*t] /.
z -> 1 + (1/4)*Exp[I*t], {t, 0, 2*Pi}]
N[4*Pi*I*E]

Out[13]=
7.105427357601002*^-15 +
34.15893689068571*I

Out[14]=
0. + 34.158936890694264*I