calculate given product & express answer in the form
- Thread starter jeef
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\(\displaystyle \L\\\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)^{10}\)
\(\displaystyle \L\\|z|=r=\sqrt{(\frac{\sqrt{3}}{2})^{2}+(\frac{1}{4})^{2}}=1\)
By DeMoivre's theorem:
\(\displaystyle \L\\1^{10}\left[cos(10\cdot\frac{\pi}{6})+i\cdot\sin(10\cdot\frac{\pi}{6})\right]=\frac{1}{2}-\frac{\sqrt{3}}{2}i\)
\(\displaystyle \L\\|z|=r=\sqrt{(\frac{\sqrt{3}}{2})^{2}+(\frac{1}{4})^{2}}=1\)
By DeMoivre's theorem:
\(\displaystyle \L\\1^{10}\left[cos(10\cdot\frac{\pi}{6})+i\cdot\sin(10\cdot\frac{\pi}{6})\right]=\frac{1}{2}-\frac{\sqrt{3}}{2}i\)
You could always use FOIL and the distrubutive property to finaly get to the 10th power.
Example: start by FOILing \(\displaystyle \LARGE (\frac {\sqrt{3}}{2} + \frac {1}{2}i)(\frac {\sqrt{3}}{2} + \frac {1}{2}i)\)
And then multiply your solution by \(\displaystyle (\frac {\sqrt{3}}{2} + \frac {1}{2}i)\) until you get to the 10th power.
galactus Let me know if this is wrong
Example: start by FOILing \(\displaystyle \LARGE (\frac {\sqrt{3}}{2} + \frac {1}{2}i)(\frac {\sqrt{3}}{2} + \frac {1}{2}i)\)
And then multiply your solution by \(\displaystyle (\frac {\sqrt{3}}{2} + \frac {1}{2}i)\) until you get to the 10th power.
galactus Let me know if this is wrong
pka
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The textbook is correct. Just use the answer given above and apply the following:
\(\displaystyle \L
\begin{array}{l}
\frac{{10\pi }}{6} \approx \frac{{ - \pi }}{3} \\
\cos \left( {\frac{{ - \pi }}{3}} \right) = \frac{1}{2}\quad \& \quad \sin \left( {\frac{{ - \pi }}{3}} \right) = \frac{{ - \sqrt 3 }}{2} \\
\end{array}\)
\(\displaystyle \L
\begin{array}{l}
\frac{{10\pi }}{6} \approx \frac{{ - \pi }}{3} \\
\cos \left( {\frac{{ - \pi }}{3}} \right) = \frac{1}{2}\quad \& \quad \sin \left( {\frac{{ - \pi }}{3}} \right) = \frac{{ - \sqrt 3 }}{2} \\
\end{array}\)
jwpaine said:
You can do it that way, but it's unwieldy. That's why we use DeMoivre's theorem.
That's all I applied. No, your answer is incorrect. It should be \(\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}i\). Which is what you get if you use what I posted.
The pi/6 comes from \(\displaystyle cos^{-1}(\frac{\sqrt{3}}{2})=\frac{\pi}{6}\)
Also, \(\displaystyle sin^{-1}(\frac{1}{2})=\frac{\pi}{6}\)
Doing it your way, Remember, because of the i, your signs will change from positive to negative here and there.
For instance, \(\displaystyle (\frac{\sqrt{3}}{2}+\frac{1}{2}i)^{4}=\frac{-1}{2}+\frac{\sqrt{3}}{2}\)
Or, of course, use pka's suggestion.