1. ## 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?

2. 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.

3. So just expand the brackets and replace z with a+bi ?