Can I Use Substitution to Prove the Residue Theorem Integral?

[MatinSAR]

Homework Statement

Prove Residue theorem.

Relevant Equations

Please see the following.

Integral 7.2 is ok. I must employ the integration technique used in 7.2 to prove that integral equation 7.1 equals zero. For n<0 we have : $$\sum_{n=- \infty}^{-2} a_n \oint (z-z_0)^ndz$$For n>0 we have : $$\sum_{n=0}^{\infty} a_n \oint (z-z_0)^ndz$$
According to Cauchy's Integral Theorem, since there are no singularities within contour C, the value of the second integral is zero. But for n<0 , $(z-z_0)^n$ isn't an analytic function so it can be unzero.

How can I show that this integral is also equal to zero? Can I show it by using the substitution $z-z_0=re^{i\theta}$ like 7.2? Or I should do sth more than this...

 

[fresh_42]

Why don't you use the given anti-derivative?

 

[MatinSAR]

Why don't you use the given anti-derivative?

Thank you for tour reply @fresh_42
Do you mean the anti-derivative term in equation 7.1?
If so, I prefer not to use that method because my professor used a different method, and I intend to use a similar approach to his. I recall that his method was similar to the approach in 7.2.

 

[fresh_42]

Thank you for tour reply @fresh_42
Do you mean the anti-derivative term in equation 7.1?

Yes.

If so, I prefer not to use that method because my professor used a different method, and I intend to use a similar approach to his. I recall that his method was similar to the approach in 7.2.

I don't think that would be a big difference, but let's see. I did the same as in 7.2. which you suggested in

\begin{align*} \oint_\gamma \dfrac{1}{z-z_0}\,dz&=\int_0^{2\pi}\dfrac{1}{re^{it}} \cdot ire^{it} \,dt=i\cdot \int_0^{2\pi}dt=2\pi i\\ \oint_\gamma \left(z-z_0\right)^p\,dz&\stackrel{(p\neq -1)}{=}\int_0^{2\pi} r^p e^{i p t} \cdot ire^{it}\,dt=ir^{p+1}\int_0^{2\pi} e^{i(p+1)t}\,dt\\ &=ir^{p+1}\left[\dfrac{e^{i(p+1)t}}{i(p+1)}\right]_0^{2\pi}=\dfrac{r^{p+1}}{p+1}(e^{2(p+1)\pi i t}-e^0)=0\,. \end{align*}

Reference: https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/

Was that what you meant?

Edit: The path here was
$\gamma(t)=z_0+re^{i t}\, , \,0\leq t\leq 2\pi$
and the path integral
\begin{align*} \int_\gamma f(z)\,dz &=…\text{ (substitution }z=\gamma(t)\, , \,dz=\gamma’\,dt) \\ …&=\int_a^b (f\circ \gamma )(t)\cdot \gamma’\,dt =\int_a^b f(\gamma (t))\cdot \gamma'(t)\,dt \end{align*}

 

[Vanadium 50]

my professor used a different method, and I intend to use a similar approach

But you aren't going to tell us what it is, hoping we hit on it by accident? That seems...inefficient and likely slow.

 

[MatinSAR]

Yes.

I don't think that would be a big difference, but let's see. I did the same as in 7.2. which you suggested in

\begin{align*} \oint_\gamma \dfrac{1}{z-z_0}\,dz&=\int_0^{2\pi}\dfrac{1}{re^{it}} \cdot ire^{it} \,dt=i\cdot \int_0^{2\pi}dt=2\pi i\\ \oint_\gamma \left(z-z_0\right)^p\,dz&\stackrel{(p\neq -1)}{=}\int_0^{2\pi} r^p e^{i p t} \cdot ire^{it}\,dt=ir^{p+1}\int_0^{2\pi} e^{i(p+1)t}\,dt\\ &=ir^{p+1}\left[\dfrac{e^{i(p+1)t}}{i(p+1)}\right]_0^{2\pi}=\dfrac{r^{p+1}}{p+1}(e^{2(p+1)\pi i t}-e^0)=0\,. \end{align*}

Reference: https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/

Was that what you meant?

It seems similar enough to my professor's method for me to use in the exam. I wrote something similar to this as well. Thank you for your time @fresh_42

But you aren't going to tell us what it is, hoping we hit on it by accident? That seems...inefficient and likely slow.

Hello. I wasn't expecting mind reading. I was just looking for a similar approach, like what was mentioned above.