Contour Integration help with understanding an example

**cooltowns** · December 30th 2010, 03:12 AM

Hi

I have attached an example from my notes for contour integration.

However I don't understand how they derive the parametrized equation $\gamma(t)=t^{2}=it$

I would appreciate if someone could please explain this to me

Thanks

**chisigma** · December 30th 2010, 03:59 AM

Take into account that is $z= x + i\ y = t^{2} + i\ t$ so that...

$\displaystyle |z|^{2} = x^{2} + y^{2} = t^{4} + t^{2}$ (1)

$\displaystyle dz = dx + i\ dy = (2\ t+ i)\ dt$ (2)

Kind regards

$\chi$ $\sigma$

**Ackbeet** · December 30th 2010, 06:00 AM

If you're wondering how they came up with $\gamma(t)=t^{2}+it,$ I would say that they wrote the example that way. There's nothing to derive with respect to the choice of contour. There's loads of freedom in deciding on contours over which to integrate, depending on the situation.

**chisigma** · December 30th 2010, 06:16 AM

It is important to note that $|z|^{2}$ isn't analytic so that the integral $\displaystyle \int_{0}^{1+i} |z|^{2}\ dz$ does depend from the path $\gamma$ connecting the point $a=0$ and $b=1+i$ ...

Kind regards

$\chi$ $\sigma$

**SammyS** · December 30th 2010, 08:57 PM

Originally Posted by cooltowns

Hi

I have attached an example from my notes for contour integration.

However I don't understand how they derive the parametrized equation $\gamma(t)=t^{2}+it$

I would appreciate if someone could please explain this to me

Thanks

I can't tell you how they get the parametrized equation:

$\gamma(t)=t^{2}=it,\ \ 0\le t\le 1\ ,$

but I can tell you the path it describes.

$\displaystyle x+i\,y = t^2+i\,t$

then $\displaystyle y = t\,,$ and $\displaystyle x = t^2\,.$

thus: $\displaystyle x = y^2\,,$

Since $\displaystyle 0\le t\le 1\ ,$ we have $\displaystyle y = \sqrt{x}\,,\ 0\le x\le 1\,.$

The path is from point (0, 0) to point (1, 1) via $\displaystyle y = \sqrt{x}\,.$

Writing that in complex terms: The path is from z=0+0i to z=1+i via $\displaystyle z = x+i\sqrt{x}\,.$

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