calculate given product & express answer in the form

[jeef]

calculate given product & express answer in the form a+bi

 

[galactus]

\(\displaystyle \L\\\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)^{10}\)

\(\displaystyle \L\\|z|=r=\sqrt{(\frac{\sqrt{3}}{2})^{2}+(\frac{1}{4})^{2}}=1\)

By DeMoivre's theorem:

\(\displaystyle \L\\1^{10}\left[cos(10\cdot\frac{\pi}{6})+i\cdot\sin(10\cdot\frac{\pi}{6})\right]=\frac{1}{2}-\frac{\sqrt{3}}{2}i\)

 

[jeef]

My book gives the answer as:

I just don't know how to arrive at it.

 

[jwpaine]

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

 

[pka]

The textbook is correct. Just use the answer given above and apply the following:
\(\displaystyle \L
\begin{array}{l}
\frac{{10\pi }}{6} \approx \frac{{ - \pi }}{3} \\
\cos \left( {\frac{{ - \pi }}{3}} \right) = \frac{1}{2}\quad \& \quad \sin \left( {\frac{{ - \pi }}{3}} \right) = \frac{{ - \sqrt 3 }}{2} \\
\end{array}\)

 

[galactus]

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

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.