solve $ z^2 -6z + 25 $ into complex conjugate I need to solve this :$$ z^2 -6z + 25 = 0$$
My book says 'complete the square' so :
1.$$ (z - 6/2)^2 -36/4 + 25 $$
2.$$ (z - 3)^2 -9 + 25 $$
3.$$ (z - 3)^2 + 16 $$  
Now how exactly does the above turn into this:
$$ 3\pm 4\imath $$
Thanks so Much
Gideon
 A: I think your confusion here stems from the fact you're working with an expression, not an equation. You can only solve equations; it makes no sense to 'solve' an expression!
Edit: You seem to have just edited your post so the first line is an equation. Best to keep it in equation form throughout however.
I'm presuming you have:
$$z^2 - 6z + 25 = 0$$
Then, completing the square:
$$(z - 6/2)^2 -36/4 + 25 = 0$$
$$(z - 3)^2 - 9 + 25 = 0$$
$$(z - 3)^2 = -16$$
$$z - 3 = \pm 4i$$
$$z = 3 \pm 4i$$
A: The square roots of -16 are $4i$ and $-4i$.

As a further hint to anybody who might encounter this sort of thing in the future, a polynomial with complex conjugate roots $\alpha$ and $\alpha^\ast$ will be of the form
$$x^2-2(\Re\alpha)x+|\alpha|^2$$
Note that $3^2+4^2=5^2=25$ and that $6=2\times 3$. Thus, once you verify that your quadratic has a negative discriminant, you can almost instantly write down the complex conjugate roots.
A: Noldorin's answer pretty much covers what I'd say, but I would note that you can proceed in working with the expression until it is factored into linear terms and read off the zeros (the solutions to the equation $(z - 3)^2 + 16=0$) from there.
Continuing from $(z - 3)^2 + 16$:
$$\begin{align}
(z - 3)^2 + 16 &= (z - 3)^2 - (-16)
\\\\
&= (z - 3)^2 - (4i)^2
\\\\
&= ((z-3)-4i)((z-3)+4i)
\\\\
&= (z-(3+4i))(z-(3-4i))
\end{align}$$
So, the zeros of the expression are $3+4i$ and $3-4i$.
A: You can also use the quadratic formula to compute the result elegantly, it goes like this :
$z^2 - 6z + 25 = 0 $
$ z = \frac { 6 \pm \sqrt{36 -100} } {2} $
$ = \frac{ 6 \pm 8i}{2} $
$=  3 \pm 4i $
