Hi
I have attached an example from my notes for contour integration.
However I don't understand how they derive the parametrized equation $\displaystyle \gamma(t)=t^{2}=it$
I would appreciate if someone could please explain this to me
Thanks
Hi
I have attached an example from my notes for contour integration.
However I don't understand how they derive the parametrized equation $\displaystyle \gamma(t)=t^{2}=it$
I would appreciate if someone could please explain this to me
Thanks
Take into account that is $\displaystyle z= x + i\ y = t^{2} + i\ t$ so that...
$\displaystyle \displaystyle |z|^{2} = x^{2} + y^{2} = t^{4} + t^{2}$ (1)
$\displaystyle \displaystyle dz = dx + i\ dy = (2\ t+ i)\ dt$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
If you're wondering how they came up with $\displaystyle \gamma(t)=t^{2}+it,$ 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.
It is important to note that $\displaystyle |z|^{2}$ isn't analytic so that the integral $\displaystyle \displaystyle \int_{0}^{1+i} |z|^{2}\ dz$ does depend from the path $\displaystyle \gamma$ connecting the point $\displaystyle a=0$ and $\displaystyle b=1+i$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
I can't tell you how they get the parametrized equation:
$\displaystyle \gamma(t)=t^{2}=it,\ \ 0\le t\le 1\ ,$
but I can tell you the path it describes.
$\displaystyle \displaystyle x+i\,y = t^2+i\,t$
then $\displaystyle \displaystyle y = t\,,$ and $\displaystyle \displaystyle x = t^2\,.$
thus: $\displaystyle \displaystyle x = y^2\,,$
Since $\displaystyle \displaystyle 0\le t\le 1\ ,$ we have $\displaystyle \displaystyle y = \sqrt{x}\,,\ 0\le x\le 1\,. $
The path is from point (0, 0) to point (1, 1) via $\displaystyle \displaystyle y = \sqrt{x}\,. $
Writing that in complex terms: The path is from z=0+0i to z=1+i via $\displaystyle \displaystyle z = x+i\sqrt{x}\,. $