Solving Answering Questions on de Moivres Theorem

[Theman123]

Homework Statement

This problem is probably quite simple for people who have covered this topic but the answer for part A is n=4k where k is any integer.

Part B answer is n= 2+4k

Look below for the working out and i will tell you what i don't understand

Homework Equations

de moivres theorem

The Attempt at a Solution

polar form is (\sqrt{2} cis \pi /4)^n

so \sqrt{2}^n [cos(n\pi /4)] +isin(n\pi /4)] (demoivres theorem)

When Sin (n\pi /4) = 0 the equation is real

n\pi /4 = 0 +k2\pi (For some reason this should be +k\pi which is why i don't get the answer.

n=8k (solving for n)

Part b is prety much the same thing but cos (n\pi /4) The question that i am asking is why is it +k\pi and not +k2\pi as that is usually the period for sin and cosine but \pi is the period for tan i know that.

 

[Saitama]

Try to find out the general solution for:-
sin\frac{n\pi}{4}=0

 

[SammyS]

Try to find out the general solution for:-
sin\frac{n\pi}{4}=0

What Pranav-Arora suggests is correct. All that's required to make (z+i)ⁿ real is for its imaginary part to be zero. It doesn't matter whether the real part is positive of if it's negative.

Similarly, to make it purely imaginary, its real part must be zero.

 

[daveb]

The question that i am asking is why is it +k\pi and not +k2\pi as that is usually the period for sin and cosine but \pi is the period for tan i know that.

Even though the period is 2\pi, sinx = 0 at both 0 and at \pi.

 

[Theman123]

so are you suggesting that both answers are correct whether i write both pi and 2pi? as sin = 0 at both, how about cos its zero at pi/2 and 3pi/2 but in the book it sais pi aswell

 

[Theman123]

Okay I've figured out out now thankyou very much!