# Solving equations in cartesian form.

• Feb 24th 2011, 02:48 AM
johnsy123
Solving equations in cartesian form.
Solve the following in cartesian form:
$\displaystyle z^4=16$
• Feb 24th 2011, 02:52 AM
Prove It
$\displaystyle \displaystyle z^4 = 16$

$\displaystyle \displaystyle z^4 - 16 = 0$

$\displaystyle \displaystyle (z^2)^2 - 4^2 = 0$

$\displaystyle \displaystyle (z^2 - 4)(z^2 + 4) = 0$

$\displaystyle \displaystyle z^2 - 4 = 0$ or $\displaystyle \displaystyle z^2 + 4 = 0$

$\displaystyle \displaystyle z^2 = 4$ or $\displaystyle \displaystyle z^2 = -4$

$\displaystyle \displaystyle z = \pm 2$ or $\displaystyle \displaystyle z = \pm 2i$.

So the solutions are $\displaystyle \displaystyle z = -2, z = 2, z = -2i, z = 2i$.
• Feb 24th 2011, 02:55 AM
mr fantastic
Quote:

Originally Posted by johnsy123
Solve the following in cartesian form:
$\displaystyle z^4=16$

Solve $\displaystyle z^4 - 16 = 0$ (I suggest that factorising is a good approach to take).

If you need more help, please post all your work and say where you get stuck.

And if solutions over the complex number field are required, you should have said so.