Complex Analysis, Integrals? 10 points for informative answer!

Hello Experts,

I have the following integral: int exp(1/(1-z)) dz/z in region : D: |z-2|+|z+2|<6

I need to understand how to get the residue at z=1.

I did series representation at z=1 of the exp(1/(1-z)): Sigma from n=0 to infinity of (-1)^n/n! * (z-1)^-n
and for 1/z : Sigma from k=0 to infinity of (-1)^k/k! * (z-1)^k

I need to multiply them but I am a little stuck there....

The residue at z=1 equal 1-e.

http://img200.imageshack.us/img200/4761/…

I mean for your part of the answer it should be 1-e, second part gives us e and together we get 2*pi*i

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To obtain the residue for f(z) = exp(1/(1 - z)) at z = 1:

Since e^t = 1 + t + t^2/2! + ..., letting t = 1/(1 - z) yields
exp(1/(1 - z)) = 1 + 1/(1 - z) + (1/2!) * 1/(1 - z)^2 + ...
....................= 1 - 1/(z - 1) + (1/2!) * 1/(z - 1)^2 - ...
....................= Σ(n = 0 to ∞) (-1)^n / [n! (z - 1)^n].

Next, we use the geometric series to expand 1/z about z = 1:
1/z = 1/[1 - (-(z - 1))]
......= 1 - (z - 1) + (z - 1)^2 - (z - 1)^3 + ...
......= Σ(k = 0 to ∞) (-1)^k (z - 1)^k

Therefore, the series for exp(1/(1 - z)) * 1/z about z = 1 is
{Σ(n = 0 to ∞) (-1)^n / [n! (z - 1)^n]} * {Σ(k = 0 to ∞) (-1)^k (z - 1)^k}

For the residue, we only want the coefficient for 1/(z - 1).
Since the first series has negative power terms, while the second series has positive power terms, we achieve this for those terms in the second series with k = n - 1.

So, the 1/(z - 1) term equals
Σ(n = 0 to ∞) {(-1)^n / [n! (z - 1)^n]} * {(-1)^(n-1) (z - 1)^(n-1)}
= Σ(n = 0 to ∞) -1 / [n! (z - 1)].

Hence, the residue at z = 1 equals
Σ(n = 0 to ∞) -1/n! = -e.

I hope this helps!