# 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)

$z^4 = -i$

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)

$z^4 = -i$

Thanx!

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

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

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

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

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