find all complex numbers, z, which satisfy the equation $\displaystyle z^2 = 6 -8i$

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

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- Mar 5th 2010, 08:37 PMmurielxfind complex numbers
find all complex numbers, z, which satisfy the equation $\displaystyle z^2 = 6 -8i$

do i begin by writing $\displaystyle z^2 - 6 - 8i = 0$ and then solve? - Mar 5th 2010, 08:48 PMProve It
Yes.

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

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

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

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

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

Now the hard part is evaluating these square roots. I'd advise converting to Polars and using DeMoivre's Theorem. - Mar 6th 2010, 05:23 AMHallsofIvy
Another way to do this is to write z= x+ yi so that $\displaystyle z^2= x^2- y^2+ 2xyi= 6- 8i$

Now solve the two equations $\displaystyle x^2- y^2= 6$ and $\displaystyle 2xy= -8$ for real x and y. - Mar 7th 2010, 07:56 PMmurielx
I used the way suggested by HallsofIvy and I have become stuck at a certain point....

$\displaystyle x^2 - y^2 = 6$

$\displaystyle 2xy = -8$

$\displaystyle y = - \frac{4}x$

.:.

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

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

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

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