- #1
Benny
- 577
- 0
Hi, I'm having trouble with two integrals.
<br /> \int\limits_{\left| z \right| = 1}^{} {\frac{{e^z }}{{z^n }}dz} <br />
<br /> \int\limits_{\left| z \right| = 1}^{} {\frac{{\cosh z}}{{z^3 }}} dz<br />
For first one I wrote
\frac{{e^z }}{{z^n }} = \frac{{e^z }}{{\left( {z - 0} \right)z^n }} = \frac{{f\left( z \right)}}{{\left( {z - 0} \right)}}.
The main theorem which was covered in the relevant theory section was Cauchy's integral formula. By inspection f(z) is not even differentiable at z = 0 because it isn't continuous there? So can I even apply the integral formula?
Ignoring that, if n <=0 then f(z) = exp(z)z^(1-n) = 0 for z = 0. So the integral is zero for n <=0. But what about for n > 0? There's a factorial in the answer. I don't really know what to do because ofr n > 0 then f(z) = exp(z)/z^(positive power) so if I plug in z = 0 I get an undefined answer.
<br /> \int\limits_{\left| z \right| = 1}^{} {\frac{{e^z }}{{z^n }}dz} <br />
<br /> \int\limits_{\left| z \right| = 1}^{} {\frac{{\cosh z}}{{z^3 }}} dz<br />
For first one I wrote
\frac{{e^z }}{{z^n }} = \frac{{e^z }}{{\left( {z - 0} \right)z^n }} = \frac{{f\left( z \right)}}{{\left( {z - 0} \right)}}.
The main theorem which was covered in the relevant theory section was Cauchy's integral formula. By inspection f(z) is not even differentiable at z = 0 because it isn't continuous there? So can I even apply the integral formula?
Ignoring that, if n <=0 then f(z) = exp(z)z^(1-n) = 0 for z = 0. So the integral is zero for n <=0. But what about for n > 0? There's a factorial in the answer. I don't really know what to do because ofr n > 0 then f(z) = exp(z)/z^(positive power) so if I plug in z = 0 I get an undefined answer.
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