- #1
Peter Alexander
- 26
- 3
Hello everyone! I'm having a bit of a problem with comprehension of the Cauchy integration formula. I might be missing some key know-how, so I'm asking for any sort of help and/or guideline on how to tackle similar problems. I thank anyone willing to take a look at my post!
Compute the given integral $$ f\left(z\right)=\frac{e^{z^{2}}}{z^{2}-6z} $$ using Cauchy integration formula for
My way of tackling problems in mathematics is to write down concise formulas and their explanations. In this particular case, I believe the relevant formulas are
To begin with, we test analyticity of ##f(z)## using Cauchy-Riemann equations. As it turns out, both equations are satisfied.
Now, this is where issues begin. To clear any ambiguities, CT refers to Cauchy Theorem (check "relevant equations" under point 1) and CIF to Cauchy integration formula (check "relevant equations" under point 2).
For both cases, I looked at singularities (singularity points). As we can see, for the given function ##f(z)## two poles exist; one at ##z_{1}=0## and the other at ##z_{2}=6##. We know that in order to use CT, the enclosed path of integration must not include any of the poles/singularities. If it does, we might be able to solve the integral using CIF.
The first case, where ##|z-2|=1## traces a unit circle with center at ##z=2##, we can see that no singularities are embraced. This way we state that according to CT, $$\int_{C:\;|z-2|=1}f(z)dz=0$$ For the second case, however, we see that a traced circle with radius 3 and center at ##z=2## does include a singularity point ##z_{1}=0##. We initially begin by writing $$f(a)=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-a}dz$$ $$\oint\frac{f(z)}{z-a}dz=2\pi if(a)$$ All we need to do now is determine what ##f(z)## and ##z-a## is. By splitting the denominator into $$\frac{e^{z^{2}}}{z(z-6)}$$ question arises; if ##f(z)=e^{z^{2}}##, then what is "##z-a##'' term?
Homework Statement
Compute the given integral $$ f\left(z\right)=\frac{e^{z^{2}}}{z^{2}-6z} $$ using Cauchy integration formula for
- ## \left|z-2\right|=1 ##
- ## \left|z-2\right|=3 ##
Homework Equations
My way of tackling problems in mathematics is to write down concise formulas and their explanations. In this particular case, I believe the relevant formulas are
- According to Cauchy theorem, if ## f(z) ## is a complex analytical function and its derivative ## f'(z) ## is continuous everywhere within the given region bounded by enclosed curve ##C##, then $$
\oint_{C} f(z)dz=0
$$ - Cauchy integration formula states that for ##f=f(z)## being an analytical function defined on a region bounded by enclosed curve ##C##, we can write $$f\left(z_{0}\right)=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-z_{0}}dz$$ provided that ##z_0## lays within such region.
- Lastly, we know that $$\oint_{C}\frac{1}{z-z_{0}}dz=2\pi i$$
The Attempt at a Solution
To begin with, we test analyticity of ##f(z)## using Cauchy-Riemann equations. As it turns out, both equations are satisfied.
Now, this is where issues begin. To clear any ambiguities, CT refers to Cauchy Theorem (check "relevant equations" under point 1) and CIF to Cauchy integration formula (check "relevant equations" under point 2).
For both cases, I looked at singularities (singularity points). As we can see, for the given function ##f(z)## two poles exist; one at ##z_{1}=0## and the other at ##z_{2}=6##. We know that in order to use CT, the enclosed path of integration must not include any of the poles/singularities. If it does, we might be able to solve the integral using CIF.
The first case, where ##|z-2|=1## traces a unit circle with center at ##z=2##, we can see that no singularities are embraced. This way we state that according to CT, $$\int_{C:\;|z-2|=1}f(z)dz=0$$ For the second case, however, we see that a traced circle with radius 3 and center at ##z=2## does include a singularity point ##z_{1}=0##. We initially begin by writing $$f(a)=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-a}dz$$ $$\oint\frac{f(z)}{z-a}dz=2\pi if(a)$$ All we need to do now is determine what ##f(z)## and ##z-a## is. By splitting the denominator into $$\frac{e^{z^{2}}}{z(z-6)}$$ question arises; if ##f(z)=e^{z^{2}}##, then what is "##z-a##'' term?