complex integral

**NaNa** · May 25th 2009, 12:44 PM

How we solve those?

**artvandalay11** · May 25th 2009, 06:15 PM

For the first one, if $n\geq0$ , then we can use Cauchy's Theorem to change the curve to a point through a homotopy and the integral over a point is zero.

For n=-1 use Cauchy's Integral formula to conclude the answer is $2\pi\imath$

For $n\leq-2$ then we have to use Cauchy's General Integral Formula which is $f^n(a)=\frac{n!}{2\pi\imath}\int_C\frac{f(z)}{(z-a)^{n+1}}$ where $f^n(a)$ is the n-th derivative of f(z) evaluated at the point a

So...

$\int_C(z-a)^ndz=\int_C\frac{1}{(z-a)^k}dz$ for k=-n and by Cauchy's Integral formula this integral is

$\frac{2\pi\imath}{(k-1)!}f^{(k-1)}(z)$ but f(z)=1 in this case, so the k-1 derivative of 1 is 0 (as k-1 is always at least 1) and anything times 0 is 0.

**artvandalay11** · May 25th 2009, 06:44 PM

$\oint_C\frac{z^3+z}{(2z+1)^3}dz=\oint_C\frac{z^3+z }{(2(z+\frac{1}{2}))^3}dz=\oint_C\frac{z^3+z}{2^3( z+\frac{1}{2})^3}dz$

$=\oint_C\frac{\frac{z^3+z}{8}}{(z+\frac{1}{2})^3}d z$

Now Cauchy's integral formula states:

$f^n(a)=\frac{n!}{2\pi\imath}\oint_C\frac{f(z)}{(z-a)^{n+1}}$ where $f^n(a)$ is the n-th derivative of f(z) evaluated at the point a

So $=\oint_C\frac{\frac{z^3+z}{8}}{(z+\frac{1}{2})^3}d z=\frac{2\pi\imath}{2!}*f ''(-\frac{1}{2})$ where f(z)= $\frac{z^3+z}{8}$

I'll leave it to you to find the second derivative, plug in z= $\ -\frac{1}{2}$ and multiply it by $\frac{2\pi\imath}{2!}$

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