Complex equation

• Mar 22nd 2010, 01:21 AM
DenOnde
Complex equation
First of all you'll have to excuse any miss-spelling an incorrect expressions, I am a swede and am not to good regarding the math grammar in english..

Now for the problem at hand:

Solve the equation and answer in polar form r(cosv + isinv)

$\displaystyle z^4 = -i$

Thanx!
• Mar 22nd 2010, 02:24 AM
Hello DenOnde

Welcome to Math Help Forum!

Your English is excellent. It's certainly much better than my Swedish!
Quote:

Originally Posted by DenOnde
First of all you'll have to excuse any miss-spelling an incorrect expressions, I am a swede and am not to good regarding the math grammar in english..

Now for the problem at hand:

Solve the equation and answer in polar form r(cosv + isinv)

$\displaystyle z^4 = -i$

Thanx!

If we write
$\displaystyle z = r(\cos\theta + i\sin\theta)$
then, using De Moivre's Theorem:
$\displaystyle z^4 = r^4(\cos\theta + i\sin\theta)^4$
$\displaystyle =r^4(\cos4\theta+i\sin4\theta)$
We now express $\displaystyle - i$ in polar form. On the Argand diagram it is represented by the point $\displaystyle (0,-1)$, so has modulus $\displaystyle 1$ and argument $\displaystyle -\pi/2$. So:
$\displaystyle -i=1\Big(\cos(-\pi/2)+i\sin(-\pi/2)\Big)$
So:
$\displaystyle z^4=-i$

$\displaystyle \Rightarrow r^4(\cos4\theta+i\sin4\theta)=\cos(-\pi/2)+i\sin(-\pi/2)$

$\displaystyle \Rightarrow \left\{\begin{array}{l} r^4\cos4\theta=-\cos(-\pi/2)\\ r^4\sin4\theta = -\sin(-\pi/2) \end{array}\right .$

$\displaystyle \Rightarrow \left\{\begin{array}{l} r=1\\ 4\theta=-\pi/2+2n\pi, n \in \mathbb{Z} \end{array}\right .$

So the values of $\displaystyle \theta$ are given by:
$\displaystyle 4\theta = -\pi/2, -\pi/2\pm2\pi,-\pi/2\pm4\pi, ...$
and its principle values (those between $\displaystyle -\pi$ and $\displaystyle \pi$) are given by:
$\displaystyle \theta = -5\pi/8, -\pi/8, 3\pi/8, 7\pi/8$