Complex Variables

• Jun 12th 2013, 03:55 PM
zhart
Complex Variables
Compute the following line integral:

γ(z²+3z+4)dz, where γ is the circle |z|=2 oriented counter-clockwise.

--Taking a Complex Variables course and I am completely lost, and there is no solution manual, or answers at the back of the book for even #s!
• Jun 12th 2013, 04:22 PM
topsquark
Re: Complex Variables
Quote:

Originally Posted by zhart
Compute the following line integral:

γ(z²+3z+4)dz, where γ is the circle |z|=2 oriented counter-clockwise.

--Taking a Complex Variables course and I am completely lost, and there is no solution manual, or answers at the back of the book for even #s!

Give the following substitution a try: $z = 2e^{i \theta}$

-Dan
• Jun 12th 2013, 04:27 PM
HallsofIvy
Re: Complex Variables
I presume that you know that a complex number can be written in the form $re^{-i\theta}$ where r is |z|, the distance from the origin ( 0) to the given point and $\theta$ is the "argument", that angle the line from the origin to z makes with the positive real axis (positive x axis). If z= a+ ib, the $r= \sqrt{a^2+ b^2}$, $\theta= arctan(b/a)$.

The point is that any point on "the circle |z|= 2" can be written $z= 2e^{i\theta}$ so $z^2+ 3z+ 4= 4e^{2i\theta}+ 6e^{I\theta}+ 4$. $dz= 2ie^{I\theta}d\theta$. The "oriented counter-clockwise" means the integral is from 0 to $2\pi$.
• Jun 12th 2013, 07:45 PM
Prove It
Re: Complex Variables
Quote:

Originally Posted by zhart
Compute the following line integral:

γ(z²+3z+4)dz, where γ is the circle |z|=2 oriented counter-clockwise.

--Taking a Complex Variables course and I am completely lost, and there is no solution manual, or answers at the back of the book for even #s!

It's been a while since I did Complex Analysis, but since the contour is closed and the function doesn't have any singular points in it, doesn't the integral have to equal 0 by Cauchy's Theorem?
• Jun 14th 2013, 09:24 AM
mnov
Re: Complex Variables
Quote:

Originally Posted by Prove It
It's been a while since I did Complex Analysis, but since the contour is closed and the function doesn't have any singular points in it, doesn't the integral have to equal 0 by Cauchy's Theorem?

I agree. It is zero.