# 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

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you cannot solve $z^2 - 6z + 25$ but you can solve $z^2 - 6z + 25 = 0$. – anon Oct 23 '10 at 13:47
thats what I meant, I edited it. – gideon Oct 23 '10 at 13:49
Well, it could have also been presented as "factor quadratic so-and-so into linear factors". – J. M. Oct 23 '10 at 13:52
+1 for showing what you had tried – Ross Millikan Oct 23 '10 at 16:47

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:

$$(z - 6/2)^2 -36/4 + 25 = 0$$

$$(z - 3)^2 - 9 + 25 = 0$$

$$(z - 3)^2 = -16$$

$$z - 3 = \pm 4i$$

$$z = 3 \pm 4i$$

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oooo ahhh so we're solving for z! It sucks having to figure out math on my own, I think I would've failed a third time if it were not for forums like this (I used to post a lot on another math forum) I just hope I get a D this time!!! I'll just be decently happy... – gideon Oct 23 '10 at 13:55
@giddy: That's right... Don't despair, just keep tackling these sorts of problems and at some point it will "click", and you'll no longer find it a pain! :) – Noldorin Oct 23 '10 at 14:57

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.

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but how does -16 get in there? The square root of (z-3)^2 + 16 should be z -3 + 4 right? – gideon Oct 23 '10 at 13:46
Nope. The sum of two squares $x^2+y^2$ factorizes differently from the perfect square trinomial $x^2+2xy+y^2=(x+y)^2$. – J. M. Oct 23 '10 at 13:48
Giddy, be very careful when taking the square root of a sum or difference of squares. The error J. M. pointed out is a "dangerous" one-- take your time understanding why what you just said was wrong... it will pay off in your future studies. – futurebird Oct 24 '10 at 0:18
haha ok yea I see... when I wrote it down in my book and then got it man I can be silly sometime! – gideon Oct 24 '10 at 16:00

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 $3-4i$.

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this does seem nice and easy since I forget how to complete the square. – gideon Oct 24 '10 at 15:45

You can also use the quadratic formula to compute the result elegantly, it goes like this :

$z^2 - 6z + 25 = 0$

$z = \frac { 6 \pm \sqrt{36 -100} } {2}$

$= \frac{ 6 \pm 8i}{2}$

$= 3 \pm 4i$

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