1. ## Complex Integration

How would I compute:

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

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

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

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

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

I have concluded

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

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

2. Originally Posted by zizou1089
How would I compute:

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

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

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

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

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

I have concluded

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

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.

3. 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.
Thats the bit Im stuck on ..

4. Originally Posted by zizou1089
Thats the bit Im stuck on ..
Do you know the quadratic formula?

5. 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.

6. 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!

7. 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??

8. 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 ?

9. Originally Posted by ques
I am new here and I am sorry that I gave out way too much!
No worries mate, welcome aboard!