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Matt's questions at Yahoo! Answers regarding the application of de Moivre's theorem

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MarkFL

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Feb 24, 2012
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Here are the questions:

Help With Trigonometry?

Can someone please help me with these three trig problems? Any help at all would be great, I'd really appreciate it. Thank you for your help.

I need to find each of the following powers and write the answer in standard form, rather than decimal form.

1. (√3 + i)5

2. (2 - 2i√3)4

3. (-2 - 2i)5
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Matt,

What we want to do is the get the complex values to have coefficients that represent the cosine (for the real part) and the sine (for the imaginary part) of some common angle, so that we can then apply de Moivre's theorem.

1.) \(\displaystyle z=\left(\sqrt{3}+i \right)^5\)

Now, if we factor out $\dfrac{1}{2}$ from the complex value, we may write:

\(\displaystyle z=2^5\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)^5\)

Now since:

\(\displaystyle \cos\left(\frac{\pi}{6} \right)=\frac{\sqrt{3}}{2},\,\sin\left(\frac{\pi}{6} \right)=\frac{1}{2}\)

we obtain:

\(\displaystyle z=2^5\left(\cos\left(\frac{\pi}{6} \right)+i\sin\left(\frac{\pi}{6} \right) \right)^5\)

Applying de Moivre's theorem, we have:

\(\displaystyle z=2^5\left(\cos\left(\frac{5\pi}{6} \right)+i\sin\left(5\frac{\pi}{6} \right) \right)=32\left(-\frac{\sqrt{3}}{2}+\frac{1}{2} \right)=-16\sqrt{3}+16i\)

2.) \(\displaystyle z=(2-2\sqrt{3}i)^4\)

Factoring out \(\displaystyle 4\) we may write:

\(\displaystyle z=4^4\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i \right)^4\)

Replacing the coefficients with trigonometric functions, we have:

\(\displaystyle z=4^4\left(\cos\left(-\frac{\pi}{3} \right)+i\sin\left(-\frac{\pi}{3} \right) \right)^4\)

Applying de Moivre's theorem, we get:

\(\displaystyle z=4^4\left(\cos\left(-\frac{4\pi}{3} \right)+i\sin\left(-\frac{4\pi}{3} \right) \right))=256\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i \right)=-128+128\sqrt{3}i\)

3.) \(\displaystyle z=(-2-2i)^5\)

Factoring out \(\displaystyle -2\sqrt{2}\) we get:

\(\displaystyle z=-2^{\frac{15}{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i \right)^5\)

Replacing the coefficients with trigonometric functions, we have:

\(\displaystyle z=-2^{\frac{15}{2}}\left(\cos\left(\frac{\pi}{4} \right)+i\sin\left(\frac{\pi}{4} \right) \right)^5\)

Applying de Moivre's theorem, we get:

\(\displaystyle z=-2^{\frac{15}{2}}\left(\cos\left(\frac{5\pi}{4} \right)+i\sin\left(\frac{5\pi}{4} \right) \right)=-2^{\frac{15}{2}}\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i \right)=128+128i\)