Complex Integration

• May 5th 2010, 12:18 PM
zizou1089
Complex Integration
How would I compute:

$I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}$

I have tried finding the poles by changing the values into complex values

where $z=e^{i\theta }$

$2cos\theta =z+\frac{1}{z}$

$2sin\theta =-i({z-\frac{1}{z}})$

I have concluded

$I=\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}$

Could someone tell me if this correct and I cant find the poles cause I cant find anyway of factorising! Thanks in advance!
• May 5th 2010, 02:18 PM
ques
Quote:

Originally Posted by zizou1089
How would I compute:

$I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}$

$=> I= - \int_{0}^{2\pi}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}$
I have tried finding the poles by changing the values into complex values

where $z=e^{i\theta }$

$2cos\theta =z+\frac{1}{z}$

$2sin\theta =-i({z-\frac{1}{z}})$

I have concluded

$I=-\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}$

Could someone tell me if this correct and I cant find the poles cause I cant find anyway of factorising! Thanks in advance!

Multiply iz with the denominator.

Simplifying will give us the denominator's expression as,

( 3 + 2i ) z^2 - 4 z + ( 2i - 3 )

Now you should be able to figure out the compex roots quite easily.(Happy)
• May 5th 2010, 03:19 PM
zizou1089
Quote:

Originally Posted by ques
Multiply iz with the denominator.

Simplifying will give us the denominator's expression as,

( 3 + 2i ) z^2 - 4 z + ( 2i - 3 ) = 0

Now you should be able to figure out the compex roots quite easily.(Happy)

Thats the bit Im stuck on (Worried)..
• May 5th 2010, 03:21 PM
Bruno J.
Quote:

Originally Posted by zizou1089
Thats the bit Im stuck on (Worried)..

Do you know the quadratic formula?
• May 5th 2010, 03:25 PM
ques
Consider ( 3 + 2i ) z^2 - 4 z + ( 2i - 3 ) = 0

For a(z^2) + bz + c = 0 the roots are

z = {-b +- (b^2 - 4ac)^(1/2) }/2a

Here z = [ 4 +- {16 - ( 2i - 3 ) (2i + 3) }]/ {2(2i + 3)}

You should be able to take it on from here.
• May 5th 2010, 03:27 PM
ques
Quote:

Originally Posted by Bruno J.
Do you know the quadratic formula?

I am new here and I am sorry that I gave out way too much!
• May 5th 2010, 03:48 PM
zizou1089
Yep got it with the quadratic formula, problem is sometimes using the formula i get a complex number within the square root, and the function cannot be factorised either, so is there a more consistent way of getting the roots??
• May 5th 2010, 03:51 PM
ques
Quote:

Originally Posted by zizou1089
Yep got it with the quadratic formula, problem is sometimes using the formula i get a complex number within the square root, and the function cannot be factorised either, so is there a more consistent way of getting the roots??

Why do you think that i^(1/5) or that ( i + 1)^(1/35) does not exist ?
• May 5th 2010, 04:15 PM
Bruno J.
Quote:

Originally Posted by ques
I am new here and I am sorry that I gave out way too much!

No worries mate, welcome aboard! (Clapping)