# find complex numbers

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• Mar 5th 2010, 09:37 PM
murielx
find complex numbers
find all complex numbers, z, which satisfy the equation $z^2 = 6 -8i$

do i begin by writing $z^2 - 6 - 8i = 0$ and then solve?
• Mar 5th 2010, 09:48 PM
Prove It
Quote:

Originally Posted by murielx
find all complex numbers, z, which satisfy the equation $z^2 = 6 -8i$

do i begin by writing $z^2 - 6 - 8i = 0$ and then solve?

Yes.

$z^2 - 6 - 8i = 0$

$z^2 - (6 + 8i) = 0$

$z^2 - (\sqrt{6 + 8i})^2 = 0$

$(z + \sqrt{6 + 8i})(z - \sqrt{6 + 8i}) = 0$

$z = -\sqrt{6 + 8i}$ or $z = \sqrt{6 + 8i}$.

Now the hard part is evaluating these square roots. I'd advise converting to Polars and using DeMoivre's Theorem.
• Mar 6th 2010, 06:23 AM
HallsofIvy
Another way to do this is to write z= x+ yi so that $z^2= x^2- y^2+ 2xyi= 6- 8i$

Now solve the two equations $x^2- y^2= 6$ and $2xy= -8$ for real x and y.
• Mar 7th 2010, 08:56 PM
murielx
I used the way suggested by HallsofIvy and I have become stuck at a certain point....

$x^2 - y^2 = 6$

$2xy = -8$

$y = - \frac{4}x$

.:.
$x^2 - (-\frac{4}x)^2 = 6$

$6 = x^2 + \frac{16}{x^2}$

so $x^4 - 6x^2 + 16 = 0$

here is where I am stuck and cant seem to find the two values of x

I know what to do after finding these values.

thanks