How to solve to the power of 35

[Atria]

should I put like -(1/2+i sqrt(3)/2)^35 here

 

[blamocur]

View attachment 29892
should I put like -(1/2+i sqrt(3)/2)^35 here

I would expect you to do better than that if I were your teacher. Are you familiar with polar coordinate representation of complex numbers (e.g., https://en.wikipedia.org/wiki/Complex_number#Polar_complex_plane) ?

 

[Steven G]

Can you explain why the minus appeared outside?

Hint: e^it = cos(t) + i sin(t) and [cos(t) + isin(t)]³⁵ = cos(t/35) + isin(t/35)

You need to find t such that cos(t) = 1/2 and cos(t) = sqrt(3)/2

 

[blamocur]

Can you explain why the minus appeared outside?

Hint: e^it = cos(t) + i sin(t) and [cos(t) + isin(t)]³⁵ = cos(t/35) + isin(t/35)

You need to find t such that cos(t) = 1/2 and cos(t) = sqrt(3)/2

This looks like a typo: cos(t/35) + isin(t/35) -- should not it be cos(35t)+i*sin(35t) ?

 

[pka]

View attachment 29892
should I put like -(1/2+i sqrt(3)/2)^35 here

With all this type of question we first need the argument.
If [imath]z= \left( {\dfrac{1}{2} -{\bf i}\dfrac{{\sqrt 3 }}{2}} \right)[/imath] then so that [imath]\arg(z)=-\dfrac{\pi}{3}~\&~|z|=1[/imath]. What is [imath]z^{35}=?[/imath]