# complex numbers equation

• September 17th 2009, 03:30 PM
Jones
complex numbers equation
Hi,

I have $(z+4i)^3 + 2(z+4i)^2 - 16 = 0$

I also know that z = -2-2i is one root, and since this is an equation with real coefficients another root is -2+2i. So i need to find the third root

Is there an easier way of solving this, other than polynomial division?
• September 17th 2009, 03:49 PM
Plato
Quote:

Originally Posted by Jones
I have $(z+4i)^3 + 2(z+4i)^2 - 16 = 0$
I also know that z = -2-2i is one root, and since this is an equation with real coefficients another root is -2+2i. So i need to find the third root

Look at this.
• September 18th 2009, 01:10 AM
Jones
So just expand the brackets and replace z with a+bi ?