what is going wrong?

what is:

\oint_C {\sqrt{z} dz}

where

C={{z\in \mathbb{C}: |z|=3}}

Using the parametrization

z=3e^{it}; t=0..2\pi

gives the solution as

-4\sqrt{3}

.

However; Using the parametrization

z=3e^{it}; t=-\pi..\pi

gives the solution as

-4\sqrt{3}i

.

Should they not be the same?

Reply 1

Nebula

Original post by user6

what is:

\oint_C {\sqrt{z} dz}

where

C={{z\in \mathbb{C}: |z|=3}}

Using the parametrization

z=3e^{it}; t=0..2\pi

gives the solution as

-4\sqrt{3}

.

However; Using the parametrization

z=3e^{it}; t=-\pi..\pi

gives the solution as

-4\sqrt{3}i

.

Should they not be the same?

\sqrt z

has two branches. When you change parameterisation you are also changing the branch (assuming you are taking

\sqrt{e^{i\theta}} = e^{i\theta/2}

).
Your second answer is correct for what is usually taken to be the principle value.

Reply 2

user6

Original post by Nebula

\sqrt z

has two branches. When you change parameterisation you are also changing the branch (assuming you are taking

\sqrt{e^{i\theta}} = e^{i\theta/2}

).
Your second answer is correct for what is usually taken to be the principle value.

I not sure I understand you? How does changing the parametrization change the branch?

Reply 3

davros

Original post by user6

I not sure I understand you? How does changing the parametrization change the branch?

Do you understand why you need a branch cut in the complex plane when trying to evaluate the integral of

\sqrt{z}

?

You won't get the same answer for your integral if you use different ranges for the parameterisation angle - that's the point :smile:

Reply 4

TeeEm

Original post by user6

I not sure I understand you? How does changing the parametrization change the branch?

have you come across branch cuts?

Reply 5

user6

Original post by davros

Do you understand why you need a branch cut in the complex plane when trying to evaluate the integral of

\sqrt{z}

?

You won't get the same answer for your integral if you use different ranges for the parameterisation angle - that's the point :smile:

So the positive real axis is a branch cut right? I not really sure why i need a branch cut though? I thought different parametrization should give the same answer?

Original post by TeeEm

have you come across branch cuts?

yes, I have never used them in practice though and I am a bit confused :frown:

Reply 6

davros

Original post by user6

So the positive real axis is a branch cut right? I not really sure why i need a branch cut though? I thought different parametrization should give the same answer?

(

As Nebula mentioned, the usual convention is to take a branch cut along the negative real axis and let theta (or t or whatever) range from

-\pi

\pi

.

You cannot allow your function to cross a branch cut - think about what happens to the value of the function if you could!

Reply 7

user6

Original post by davros

As Nebula mentioned, the usual convention is to take a branch cut along the negative real axis and let theta (or t or whatever) range from

-\pi

\pi

.

You cannot allow your function to cross a branch cut - think about what happens to the value of the function if you could!

This is what is confusing me? How do I decide on a branch cut e.g. why the negative real axis and not the positive real axis. I know this is a simple case - this is the easiest example I could think of where this problem is present - more generally how do I decide on a branch cut?

Reply 8

TeeEm

Original post by user6

So the positive real axis is a branch cut right? I not really sure why i need a branch cut though? I thought different parametrization should give the same answer?

yes, I have never used them in practice though and I am a bit confused :frown:

when you have fractional powers or logs you have a branch cut starting from the branch point (here z=0) all the way to infinity
so different parameterizations produce different answers depending where the cut is