Circle radius 2 complex integration

[Dustinsfl]

Gamma is a circle of radius 2 oriented counterclockwise.

$$
\int_{\gamma}\frac{dz}{z^2+1} = \int_{\gamma} = \frac{i}{2}\left[\int_{\gamma}\frac{1}{z+i}dz-\int_{\gamma}\frac{1}{z-i}dz\right]
$$

$\gamma(t) = 2e^{it}, \ \ \gamma'(t) = 2ie^{it}$

$$
\int_{\gamma}\frac{2ie^{it}}{2e^{it}+i}dz
$$

Now what can I do from here (just looking at the first integral)?

 

[chisigma]

Using the residue theorem You easily find that is $\displaystyle \int_{\gamma} \frac{d z}{1+z^{2}}=0$...

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

Using the residue theorem You easily find that is $\displaystyle \int_{\gamma} \frac{d z}{1+z^{2}}=0$...

Kind regards

$\chi$ $\sigma$

I know it is zero but I am not trying to obtain the result that way. I should get each integral is $2\pi i$ and when subtracted they are 0.

 

[Chris L T521]

Gamma is a circle of radius 2 oriented counterclockwise.

$$
\int_{\gamma}\frac{dz}{z^2+1} = \int_{\gamma} = \frac{i}{2}\left[\int_{\gamma}\frac{1}{z+i}dz-\int_{\gamma}\frac{1}{z-i}dz\right]
$$

$\gamma(t) = 2e^{it}, \ \ \gamma'(t) = 2ie^{it}$

$$
\int_{\gamma}\frac{2ie^{it}}{2e^{it}+i}dz
$$

Now what can I do from here (just looking at the first integral)?

First off, the integral you need to evaluate is $\displaystyle\int_{\gamma}\frac{dz}{z+i} = \int_0^{2\pi} \frac{2ie^{it}}{2e^{it}+i}\,dt$, given that $\gamma(t) = 2e^{it}$, $t\in[0,2\pi]$.

To integrate that, you need to make the substitution $u=2e^{it}+i\implies\,du = 2ie^{it}\,dt$ (also note that the limits of integration change as well) such that the integral becomes $\displaystyle\int_{2+i}^{2+i}\frac{du}{u} = 0$.

The other integral is computed in a similar fashion.

I hope this helps!

 

[Dustinsfl]

First off, the integral you need to evaluate is $\displaystyle\int_{\gamma}\frac{dz}{z+i} = \int_0^{2\pi} \frac{2ie^{it}}{2e^{it}+i}\,dt$, given that $\gamma(t) = 2e^{it}$, $t\in[0,2\pi]$.

To integrate that, you need to make the substitution $u=2e^{it}+i\implies\,du = 2ie^{it}\,dt$ (also note that the limits of integration change as well) such that the integral becomes $\displaystyle\int_{2+i}^{2+i}\frac{du}{u} = 0$.

The other integral is computed in a similar fashion.

I hope this helps!

our prof said there is no u sub for complex analysis because every closed path would be zero when that isn't always the case.

 

[Dustinsfl]

Is this valid:

$\int_0^{2\pi}\frac{dz}{z^2+1} = \left[\tan^{-1}(z)\right|_0^{2\pi}=0$

 

[Also sprach Zarathustra]

Is this valid:

$\int_0^{2\pi}\frac{dz}{z^2+1} = \left[\tan^{-1}(z)\right|_0^{2\pi}=0$

In real analysis, above integral equals $\arctan(2\pi)$ (the area most exist, can't be $0$)

 

[Dustinsfl]

In real analysis, above integral equals $\arctan(2\pi)$ (the area most exist, can't be $0$)

This is a line integral in complex analysis and I know the answer to this integral is 0.
The question is how to show it without using Cauchy Integral Formula or Residue Theory.

 

[Prove It]

our prof said there is no u sub for complex analysis because every closed path would be zero when that isn't always the case.

Once the function is in terms of real variables (in this case t), i can be treated as a constant and u substitutions are then valid...

 

[Dustinsfl]

Once the function is in terms of real variables (in this case t), i can be treated as a constant and u substitutions are then valid...

I think I understand what my professor means. If we substitute,

$\int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt$

The numerator is the derivative of the denominator so substitutions is viable here as well.

$\int_1^1\frac{du}{u}=0\neq 2\pi i$.

Also, $\int_0^{2\pi}\frac{dz}{z+i} = 2\pi i$ by Cauchy's Integral Formula. However, the intend is not to use that. So how can you make a substitution that says the integral is 0 when we know that is simply not true.

---------- Post added at 10:58 AM ---------- Previous post was at 10:06 AM ----------

Here is what I am thinking now.

$\displaystyle\int_0^{2\pi}\frac{dz}{z+i} = \int_0^{2\pi}\frac{dz}{x+i(y+1)}$

Let $w=y+1$. Then
$\displaystyle\int_0^{2\pi}\frac{dz}{x+iw} =\int_0^{2\pi}\frac{dz}{z_0}$

Could this work?

 

[Opalg]

If we substitute,

$\int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt$

If you then cancel $e^{it}$ top and bottom in that fraction, you get $\int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt = \int_0^{2\pi}i\,dt = \bigl[it\bigr]_0^{2\pi} = 2\pi i,$ as required.

 

[Dustinsfl]

If you then cancel $e^{it}$ top and bottom in that fraction, you get $\int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt = \int_0^{2\pi}i\,dt = \bigl[it\bigr]_0^{2\pi} = 2\pi i,$ as required.

I understand that. That wasn't the point of that example.