Finding poles for cauchy's residue theorem.

[stephen cripps]

Homework Statement

In order to use cauchy's residue theorem for a question, I need to put
$f(x)=\frac{z^{1/2}}{1+\sqrt{2}z+z^2}$
Into the form
$f(x)=\frac{\phi(z)}{(z-z_0)^m}$.
Where I can have multiple forms of
${(z-z_0)^m}$
on the denominator, e.g
$f(x)=\frac{z^{1/2}}{(z+1)(z+3)^3}$
I just need to find what values of z will take it to zero

Homework Equations

The Attempt at a Solution

The closest I have gotten is
$f(x)=\frac{Z^{1/2}}{(z+\frac{1}{\sqrt{2}})^2+\frac{1}{2}}$.
But I need to get rid of that half on the end of the denominator (I think) in order to get to the form I want.
$f(x)=\frac{Z^{1/2}}{(z+\frac{\sqrt{2}}{2}+i)(z+\frac{\sqrt{2}}{2}-i)-\frac{1}{2}}$.
was another close attempt. Can anyone help me find right factorisation?

 

[gneill]

Use the quadratic formula on the denominator to extract its roots.

 

[vela]

From where you got, you can do this:
$$\left(z + \frac{1}{\sqrt 2}\right)^2+\frac 12 = \left(z + \frac{1}{\sqrt 2}\right)^2- \left(\frac 1{\sqrt 2}i\right)^2$$ and then factor the difference of squares the usual way.

 

[stephen cripps]

From where you got, you can do this:
$$\left(z + \frac{1}{\sqrt 2}\right)^2+\frac 12 = \left(z + \frac{1}{\sqrt 2}\right)^2- \left(\frac 1{\sqrt 2}i\right)^2$$ and then factor the difference of squares the usual way.

Awesome this was exactly what I needed. THANKS!