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?
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.
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