I need help to Simplify in Polar Form please

[val1]

Simplify

\(\displaystyle \L

\frac{{\left( {\frac{{\cos \pi }}{6} + \iota \frac{{\sin \pi }}{6}} \right)^3 }}{{\left( {\frac{{\cos \pi }}{6} - \iota \frac{{\sin \pi }}{6}} \right)^5 }}\)


Any help in simplifying this would be really appreciated.

 

[soroban]

Hello, val1!

Are you sure you typed it correctly?
\(\displaystyle \;\;\)As written, it's a trivial problem . . .

Simplify: \(\displaystyle \;\L\frac{{\left( {\frac{{\cos \pi }}{6} + i\frac{{\sin \pi }}{6}} \right)^3 }}{{\left( {\frac{{\cos \pi }}{6} - i\frac{{\sin \pi }}{6}} \right)^5 }}\)

We have: \(\displaystyle \L\:\frac{\left(\frac{-1}{6}\,+\,i\cdot\frac{0}{6}\right)^3}{\left(\frac{-1}{6}\,-\,i\cdot\frac{0}{6}\right)^5}\;=\;\frac{\left(-\frac{1}{6}\right)^3}{\left(-\frac{1}{6}\right)^5} \;= \;\left(-\frac{1}{6}\right)^{-2} \;= \;36\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

More interesting is: \(\displaystyle \L\;\frac{\left(\cos\frac{\pi}{6}\,+\,i\cdot\sin\frac{\pi}{6}\right)^3}{\left(\cos\frac{\pi}{6}\,-\,i\cdot\sin\frac{\pi}{6}\right)^5}\)

DeMoivre's Theorem: \(\displaystyle \L\:\frac{\cos\left(3\cdot\frac{\pi}{6}\right)\,+\,i\cdot\sin\left(3\cdot\frac{\pi}{6}\right)}{\cos\left(5\cdot\frac{\pi}{6}\right)\,-\,i\cdot\sin\left(5\cdot\frac{\pi}{6}\right)}\;=\;\frac{\cos\frac{\pi}{2}\,+\,i\cdot\sin\frac{\pi}{2}}{\cos\frac{5\pi}{6}\,-\,i\cdot\sin\frac{5\pi}{6}}\;=\;\frac{0\,+\,i\cdot1}{\frac{\sqrt{3}}{2}\,-\,i\cdot\frac{1}{2}}\;=\;\frac{2i}{\sqrt{3}\,-\,i}\)

Rationalize: \(\displaystyle \L\:\frac{2i}{\sqrt{3}\,-\,1}\cdot\frac{\sqrt{3}\,+\,1}{\sqrt{3}\,+\,1} \;= \;\frac{2i(\sqrt{3}\,+\,1)}{4}\;=\;\frac{i(\sqrt{3}\,+\,i)}{2} \;= \;\frac{-1\,+\,i\sqrt{3}}{2}\)

 

[pka]

If the problem is indeed \(\displaystyle \L
\frac{{\left( {\cos \left( {\frac{\pi }{6}} \right) + i\sin \left( {\frac{\pi }{6}} \right)} \right)^3 }}{{\left( {\cos \left( {\frac{\pi }{6}} \right) + i\sin \left( {\frac{\pi }{6}} \right)} \right)^5 }}\)
then using the basic laws of powers and division we get:

\(\displaystyle \L
\left( {\cos \left( {\frac{{ - 2\pi }}{6}} \right) + i\sin \left( {\frac{{ - 2\pi }}{6}} \right)} \right) = \frac{1}{2} - i\frac{{\sqrt 3 }}{2}\).

There are some minor sign errors made above.

 

[val1]

You were completely right to spot that I had written the problem incorrectly.

Thank you Soroban, for your clear and helpful explanation.