Complex Numbers and Trigonometry

[adeel17]

Can anybody let me know how the step is taken at fifth line where is (cos n alpha + i sin n alpha) gone?

 

[pka]

Can anybody let me know how the step is taken at fifth line where is (cos n alpha + i sin n alpha) gone?

If I understand correctly which is the fifth line: \(\rho^n\left(cos(n\alpha)+i\sin(n\alpha)\right)=r\left(\cos(\theta)+i\sin(\theta)\right)\)
That is just a rewrite of the fourth line \(z^n=a\) where \(z=cos(\alpha)+isin(\alpha)\) & \(a=r\left(\cos(\theta)+i\sin(\theta)\right)\).

 

[Dr.Peterson]

Can anybody let me know how the step is taken at fifth line where is (cos n alpha + i sin n alpha) gone?

If you mean going from [MATH]r(\cos(\theta) + i \sin(\theta))[/MATH] to [MATH]r[\cos(2k\pi + \theta) + i \sin(2k\pi + \theta)][/MATH], they just replaced [MATH]\theta[/MATH] with [MATH]2k\pi + \theta[/MATH], because adding [MATH]2k\pi[/MATH] to the argument of a sine or cosine does not change its value -- that is, these functions have period [MATH]2\pi[/MATH]. The important idea is that because a complex number can be represented by many angles, there are n nth roots.

 

[Maths_Mike]

Obviously Cos(na) = Cos(theta) now solve and use general formula for getting all solutions.
They just did that bit first