# demoivre theorem

• Nov 14th 2006, 11:41 PM
SHORTY
demoivre theorem
write in rectangular form

(square root 2-i) 4th power
• Nov 15th 2006, 04:41 AM
topsquark
Quote:

Originally Posted by SHORTY
write in rectangular form

(divied sign 2-i) 4th power

Do you mean $\frac{1}{(2-i)^4}$?

We need to rationalize this before we can work with it, so:
$\frac{1}{(2-i)^4} = \frac{1}{(2-i)^4} \cdot \frac{(2+i)^4}{(2+i)^4}$

= $\frac{(2+i)^4}{(2-i)^4(2+i)^4} = \frac{(2+i)^4}{((2-i)(2+i))^4}$

= $\frac{(2+i)^4}{(4 - i^2)^4} = \frac{(2+i)^4}{(4+1)^5}$

= $\frac{(2+i)^4}{5^4} = \frac{(2+i)^4}{625}$

So we need to know what $(2+i)^4$ is.

Note that $2 + i = r cos \theta + ir sin \theta$.

So
$r cos \theta = 2$
$r sin \theta = 1$

Dividing the two equations gives:
$tan \theta = \frac{1}{2}$
So
$\theta \approx 0.463648$ rad.

Squaring the two equations and adding gives:
$r^2 sin^2 \theta + r^2 cos^2 \theta = 1^2 + 2^2$

$r^2(sin^2 \theta + cos^2 \theta ) = 5$

$r^2 = 5$

$r = \sqrt{5}$.

Thus $2 + i \approx \sqrt{5}cos(0.463648) + i \sqrt{5}sin(0.463648)$.

By DeMoivre's theorem:
$(cos\theta + i sin\theta)^n = cos(n\theta) + i sin(n\theta)$

So
$(2 + i)^4 \approx (\sqrt{5}cos(0.463648) + i \sqrt{5}sin(0.463648))^4$

= $(\sqrt{5})^4 (cos(4 \cdot 0.463648) + i sin(4 \cdot 0.463648))$

= $25(cos(1.85459) + i sin(0.85459))$

= $25(-0.28 + 0.96i)$ <-- These are apparently exact, though I don't know how to prove it.

So finally:
$\frac{1}{(2-i)^4} = \frac{(2+i)^4}{625} = \frac{25(-0.28 + 0.96i)}{625}$

= $\frac{-0.28 + 0.96i}{25}$

-Dan