Expressing a complex number in cartesian form

Im stuck on a textbook question that has no previous examples. Express $\displaystyle \left(\frac{\sqrt{3}-i}{2}\right)^{101}$ in cartesian form.

OK. So i know De moivre's Theorem but not sure if i can use it here (and thats going away from Cartesian form too)

But could I say that here the form r(cos x + isin x) can be deduced because $\displaystyle \frac{\sqrt{3}}{2} $ is $\displaystyle cos\frac{\pi}{6}$ and then deduce sin.....

But not sure where to go from here!...

Re: Expressing a complex number in cartesian form

Quote:

Originally Posted by

**FelixHelix** Im stuck on a textbook question that has no previous examples. Express $\displaystyle \left(\frac{\sqrt{3}-i}{2}\right)^{101}$ in cartesian form.

OK. So i know De moivre's Theorem but not sure if i can use it here (and thats going away from Cartesian form too)

But could I say that here the form r(cos x + isin x) can be deduced because $\displaystyle \frac{\sqrt{3}}{2} $ is $\displaystyle cos\frac{\pi}{6}$ and then deduce sin.....

But not sure where to go from here!...

$\displaystyle z = \left(\frac{\sqrt 3}{2}-\frac{1}{2}i\right)^{101}=\left(\cos \frac{11 \pi}{6}+i\sin\frac{11 \pi}{6}\right)^{101} $

$\displaystyle z= \cos \frac{1111 \pi}{6}+i\sin\frac{1111 \pi}{6} = \cos \frac{7 \pi}{6}+i\sin\frac{7 \pi}{6} $

$\displaystyle z = -\frac{\sqrt 3}{2}-\frac{1}{2}i $

Re: Expressing a complex number in cartesian form

Quote:

Originally Posted by

**FelixHelix** Im stuck on a textbook question that has no previous examples. Express $\displaystyle \left(\frac{\sqrt{3}-i}{2}\right)^{101}$ in cartesian form....

$\displaystyle \frac{\sqrt{3}-i}{2}=2\exp\left(\frac{-\pi}{6}\right)$

Now apply De moivre's Theorem.

Re: Expressing a complex number in cartesian form

I'm confused by the workings here. Could you explain how you get the original argument here?

Re: Expressing a complex number in cartesian form

HI Plato, Im confused, how come your workings are different to Princeps? and how do you get r = 2 in your polar form here?

Re: Expressing a complex number in cartesian form

Quote:

Originally Posted by

**FelixHelix** HI Plato, Im confused, how come your workings are different to Princeps? and how do you get r = 2 in your polar form here?

They are not different; they are equivalent. You should know that.

I just prefer the principal argument, $\displaystyle -\pi<\theta\le\pi$.