DeMoivre's Theorem

**dashreeve** · July 30th 2008, 08:00 AM

I am using DeMoivre's Theorem to find the 6th power of the complex number (2+3i). I start by putting the complex number in trig form, but am having trouble with the angle. I have r=sqrt(13), and tan0=3/2. The solution guide has an angle for this tan, but I can not determine how to get it. Also the angle they get is then used as a reference angle. I thought I could use tan(3/2) and then use that value with arctan to find the angle but it is not working.

**Soroban** · July 30th 2008, 09:06 AM

Hello, dashreeve!

Use DeMoivre's Theorem to find: . $(2 + 3i)^6$

I start by putting the complex number in trig form, but am having trouble with the angle.
I have: . $r\:=\:\sqrt{13},\;\;\tan\theta \:=\:\frac{3}{2}$ . . . . Good!

We have: . $z \;=\;2 + 3i \;=\;r(\cos\theta + i\sin\theta)$

. . And: . $\tan\theta \,=\,\frac{3}{2}\quad\Rightarrow\quad \sin\theta \,=\,\frac{3}{\sqrt{13}} \quad\Rightarrow\quad \cos\theta \,=\,\frac{2}{\sqrt{13}}$ .[1]

. . Then: . $z^6 \;=\;(\sqrt{13})^6\,\left(\cos\theta + i\sin\theta\right)^6 \;=\;13^3\,(\cos6\theta + i\sin6\theta)$

We can use DeMoivre's Theorem to derive the following identities:

. . $\begin{array}{ccc}\cos6\theta &=& \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta \\ \sin6\theta &=& 6\cos^5\theta\sin\theta -20\cos^3\theta\sin^3\theta + 6\cos\theta\sin^5\theta \end{array}$

Hence: . $z^6 \;=\;13^3\,\bigg[\left(\cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta\right)$
. . . . . . . . . . . . . . $+ i\left(6\cos^5\theta\sin\theta -20\cos^3\theta\sin^3\theta + 6\cos\theta\sin^5\theta\right)\bigg]$

Then substitute the values from [1].

**dashreeve** · July 30th 2008, 09:53 AM

Thanks for the help!!

Thread: DeMoivre's Theorem

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