# Complex Numbers Quadratic. Sorry if my explanation is a bit long.

1. May 5, 2009

### Sensayshun

1. The problem statement, all variables and given/known data

$$2iz^2 - (3-8i)z -6 + 7i$$

2. Relevant equations

$$z = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a}$$

3. The attempt at a solution

Right, here goes.....

a = 2i
b = -3 + 8i (is this correct? Or would it be better to leave it as 3-8i?)
c = (-6 + 7i)

so using these:

z = $$\frac{3-8i) +/- \sqrt{(-55-48i) - (4(2i(-6 + 7i))}}{4i}$$

I'll just work with the top line for now to make writing it easier.
But that last section under the square root, simplifies to:

$$\sqrt{(-55 - 48i) - (48i + 56)}$$
which is then:
$$\sqrt{-111 - 96i}$$

THIS IS WHERE I THINK I START TO GET STUCK

Do I then take the complex conjugate of the bottom to give:

$$\frac{(3 - 8i)(-4i)}{(4i)(-4i)} +/- \frac{\sqrt{111 - 96i}}{4i}$$

multiplying this out gives:

$$\frac{-3}{4}i - 2 +/- \frac{\sqrt{111 - 96i}}{4i}$$

and then i'm not sure where to go from here, or if I've headed in the correct direction

Thanks for any help given.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: May 5, 2009
2. May 5, 2009

### HallsofIvy

Staff Emeritus
Do you mean $z^3- (3-8i)z- 6+7i= 0$?

2. Relevant equations

$$z = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a}$$

3. The attempt at a solution

Right, here goes.....

a = 2i
b = -3 + 8i (is this correct? Or would it be better to leave it as 3-8i?)[/quote]
Yes, that is correct. You could leave it as -(3- 8i) but it is not -3+8i.

4ac= 4(2i)(-6+7i)= 8i(-6+7i)= -48i- 56 so -4ac= 48i+ 56 not -(48i+ 56)

$$\sqrt{-55- 48i+ 48i+ 56}= \sqrt{1}= 1$$
That should make it much easier!

3. May 5, 2009

### Staff: Mentor

Under the problem statement, you give this:
What exactly are you trying to do? This is not an equation, so there is nothing to solve. It is an expression, which you can factor or otherwise rewrite in some different (but equal) form.

4. May 5, 2009

### Sensayshun

To mark, sorry it's meant to be = 0 on the end. Certain bits I couldn't get to format correctly, and I thought it'd be clear(ish).

HallsOfIvy - Thanks ever so much, that might make things far simpler!
Me and my friends were sat around trying to do this and we thought we'd have gone wrong with the signs somewhere, but we couldn't see it, so assumed it was fine. Someone did end up with -1, but noone got just 1. :P I'll work through it now with what you've shown.

Thanks very much, greatly appreciated :)