# DeMoivre's Theorem

• December 25th 2010, 06:22 AM
ironz
DeMoivre's Theorem
I need to find the real and imaginary values of z^5, when z = 2-2i

I got r = 2 sqrt(2) and theta = -pi/4

However now I am stuck. I tried to put it into the form r^k e^(i k theta)

and got (2 rt(2)) ^5 * e^(-5pi/4) but then I don't know what to do to find the real and imaginary parts.

What should I do?

Thank you :)
• December 25th 2010, 06:40 AM
Plato
$\text{Re} \left( {\left[ {x + yi} \right]^5 } \right) = \sum\limits_{k = 0}^2 {\left( { - 1} \right)^k \dbinom{ 5}{2k} x^{5 - 2k} y^{2k} }$
• December 25th 2010, 07:15 AM
Soroban
Hello, ironz!

Quote:

$\text{Find the real and imaginary components of }\,(2 - 2i)^5$

Write $2 - 2i$ in polar form.
. . We have: . $x = 2,\;y =2,\;r = 2\sqrt{2},\;\theta = \text{-}\dfrac{\pi}{4}$

Hence: . $2 - 2i \:=\:2\sqrt{2}\left(\cos\frac{\pi}{4} - \sin\frac{\pi}{4}\right)$

Then: . $(2-2i)^5 \;=\;\left(2\sqrt{2}\right)^5\,\left(\cos\frac {5\pi}{4} - \sin\frac{5\pi}{4}\right)$

. . . . . . . . . . . . $=\;2^5(\sqrt{2})^5\left(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right)$

. . . . . . . . . . . . $=\;32(4\sqrt{2})\left(\dfrac{-1+i}{\sqrt{2}}\right)$

. . . . . . . . . . . . $=\;128(-1 + i)$

. . . . . . . . . . . . $=\;\underbrace{-128}_{\mathbb{R}} + \underbrace{128}_{\mathbb{I}}i$

• December 25th 2010, 10:02 AM
FernandoRevilla
Quote:

Originally Posted by ironz
I need to find the real and imaginary values of z^5, when z = 2-2i

An alternative: taking into account that $(1-i)^2=1-1-2i=-2i$:

$(2-2i)^5=2^5(-2i)^2(1-i)=-2^7(1-i)=-128+128i$

In general this is the shortest way for computing $(x+yi)^n$ when $y=x$ or $y=-x$ .

Fernando Revilla
• December 25th 2010, 10:49 AM
Quote:

Originally Posted by ironz
I need to find the real and imaginary values of z^5, when z = 2-2i

I got r = 2 sqrt(2) and theta = -pi/4

However now I am stuck. I tried to put it into the form r^k e^(i k theta)

and got (2 rt(2)) ^5 * e^(-5pi/4) but then I don't know what to do to find the real and imaginary parts.

What should I do?

Thank you :)

You reached

$\displaystyle\ z^5=\left[2\sqrt{2}\right]^5e^{-i\frac{5{\pi}}{4}}$

To regenerate the real and imaginary parts from this...

$\displaystyle\ e^{-i\frac{5{\pi}}{4}}=cos\left(-\frac{5{\pi}}{4}\right)+isin\left(-\frac{5{\pi}}{4}\right)$

$\displaystyle\ -\frac{5{\pi}}{4}=2{\pi}-\frac{5{\pi}}{4}=\frac{3{\pi}}{4}$

$\displaystyle\ cos\left(\frac{3{\pi}}{4}\right)=-cos\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}$

$\displaystyle\ sin\left(\frac{3{\pi}}{4}\right)=sin\left(\frac{\p i}{4}\right)=\frac{1}{\sqrt{2}}$

Then

$\displaystyle\ z^5=\frac{-2^5\left(\sqrt{2}^5\right)}{\sqrt{2}}+\frac{i2^5\l eft(\sqrt{2}^5\right)}{\sqrt{2}}$
• December 26th 2010, 12:56 AM
HallsofIvy
Don't let those square roots of 2 in Archie Meade's response fool you. When you actually do the calculation, they will cancel out and give the same answer FernandoRevilla gives.
• December 26th 2010, 04:19 AM
ironz
I understand. I think I read the question wrong and tried to solve for all the roots instead of just the z^5, also, I didn't know how to simplify it properly. Thank you for your help :)
• December 26th 2010, 04:30 AM
Having found $z^5$
$z=2-2i$ is one of the 5 "fifth roots" of $z^5$