# Factorising Imaginary numbers

• May 3rd 2010, 02:58 PM
zizou1089
Factorising Imaginary numbers
I am having trouble finding the poles z1 and z2 of f defined:

$f(z)=\frac{1}{2z^{2}+(6-4i)z-12i}$

Im pretty sure I have to factorise the denominator, but I keep getting the wrong answer which includes imaginary numbers, is it supposed to?
• May 3rd 2010, 06:35 PM
You get 2(z+3)(z-2i) I assume?
• May 4th 2010, 06:35 AM
zizou1089
That makes sense but how would you factorise when there's an imaginary number involved? Is there a formula of some sort??
• May 4th 2010, 06:51 AM
Tekken
Quote:

Originally Posted by zizou1089
That makes sense but how would you factorise when there's an imaginary number involved? Is there a formula of some sort??

You could use the "-b" formula if you want, i trust your familar enough with that.

Assuming maddas' solution is correct $f(z)$ has poles at $2(z+3)(z-2i) = 0$ so pole at $z = -3$ and $z = 2i$

Yes it is common that in these types of problems that the poles contain imaginary numbers
• May 4th 2010, 07:08 AM
zizou1089
I tried using the '-b' formula but I get the square root of an imaginary number which I don't think should be the case..
• May 4th 2010, 09:07 AM
zizou1089
For example how would I factorise:

$(i+\frac{3}{2})z^2-2z+(i-\frac{3}{2})$