What is the Result of Raising a Complex Number to a Power?

[Petrus]

Hello MHB,
calculate \(\displaystyle \left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{100}\) in the form \(\displaystyle a+ib\)

progress:
I start to calculate argument and get it to \(\displaystyle r=1\) (argument)
then \(\displaystyle \cos\theta=\frac{1}{2} \ sin\theta=\frac{\sqrt{3}}{2}\) we se it's in first quadrant( where x and y is positive)
\(\displaystyle 1*e^{i\frac{100\pi}{3}}\)
notice that we can always take away 2pi so we can simplify that to
\(\displaystyle 1*e^{i\frac{\pi}{3}}\)
\(\displaystyle 1*e^{i\frac{\pi}{3}}=\cos(\frac{\pi}{3}) + i \sin (\frac{\pi}{3}) = \frac{1}{2}+i\frac{\sqrt{3}}{2}\)
but the facit says \(\displaystyle \frac{-1}{2}-i\frac{\sqrt{3}}{2}\)

Regards,

 

[I like Serena]

Hello MHB,
calculate \(\displaystyle \left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{100}\) in the form \(\displaystyle a+ib\)

progress:
I start to calculate argument and get it to \(\displaystyle r=1\) (argument)
then \(\displaystyle \cos\theta=\frac{1}{2} \ sin\theta=\frac{\sqrt{3}}{2}\) we se it's in first quadrant( where x and y is positive)
\(\displaystyle 1*e^{i\frac{100\pi}{3}}\)
notice that we can always take away 2pi so we can simplify that to
\(\displaystyle 1*e^{i\frac{\pi}{3}}\)
\(\displaystyle 1*e^{i\frac{\pi}{3}}=\cos(\frac{\pi}{3}) + i \sin (\frac{\pi}{3}) = \frac{1}{2}+i\frac{\sqrt{3}}{2}\)
but the facit says \(\displaystyle \frac{-1}{2}-i\frac{\sqrt{3}}{2}\)

Regards,

Hey Petrus!

What is \(\displaystyle \frac {100\pi}{3} \pmod{2\pi}\)?

 

[Petrus]

Hey Petrus!

What is \(\displaystyle \frac {100\pi}{3} \pmod{2\pi}\)?

\(\displaystyle \frac{4}{3}\)

- - - Updated - - -

Thanks I like Serena I see what I did wrong :)

Regards,

 

[I like Serena]

\(\displaystyle \frac{4}{3}\)

That should be \(\displaystyle \frac{4\pi}{3}\). (Yeah, I know, I'm a nitpicker.)

So what's \(\displaystyle \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\)?

 

[Petrus]

(Yeah, I know, I'm a nitpicker

nitpicker or not, I am grateful for the fast responed!

Regards,

 

[MarkFL]

If I were to work the problem, I would write:

\(\displaystyle \left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{100}= \left(\cos\left(\frac{\pi}{3}+2k\pi \right)+i\sin\left(\frac{\pi}{3}+2k\pi \right) \right)^{100}\)

Now applying de Moivre's theorem we have:

\(\displaystyle \cos\left(\frac{100\pi}{3}+200k\pi \right)+i\sin\left(\frac{100\pi}{3}+200k\pi \right)=\cos\left(\frac{4\pi}{3}+232k\pi \right)+i\sin\left(\frac{4\pi}{3}+232k\pi \right)= -\left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)\)