Why is \(\theta = n(\tan^{-1}(\frac{b}{a}))\) in complex exponentials?

[Gregg]

| a + bi | = \sqrt{(a)^2+(b)^2}
\theta = tan^{-1}(\frac{b}{a})

|(a + bi)^n| = (\sqrt{(a)^2+(b)^2})^n
\theta = n(tan^{-1}(\frac{b}{a}))

(a+bi) = (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))

Why is \theta = n(tan^{-1}(\frac{b}{a}) ? For example when I have

5 + i

\theta = tan^{-1}(\frac{1}{5}) = 0.1974...

(5+i)^2 = 25 + 10i -1

(5+i)^2 = 24 + 10i

\alpha = tan^{-1}(\frac{10}{24}) = 0.3948...

2(0.1974) =0.3948

Why is it that the exponent of the vector can be used to get the angle of the resultant by simply multiplying it with the tan function? Also in the first part of that:

(\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))

Just to make sure, it is only

isin(n(tan^{-1}(\frac{b}{a}))))

and not simply

(\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + sin(n(tan^{-1}(\frac{b}{a}))))

because one of the components is complex?

 

[HallsofIvy]

you are talk about putting complex numbers in "polar form". If you represent a+ bi as the point (a, b) in Cartesian coordinates and the polar form is (r, \theta) (That is, distance from (0,0) to (a,b) is r and the line from (0,0) to (a,b) is \theta, it is easy to see that a= rcos(\theta) and b= rsin(\theta) so that a+ bi= r (cos(\theta)+ i sin(\theta)). Then [/itex](a+bi)^2= r^2(cos^2(\theta)-sin^2(\theta)+ i(2sin(\theta)cos(\theta)))[/itex] and recognise that cos^2(\theta)- sin^2(\theta)= cos(2\theta) and [/itex]2cos(\theta)sin(\theta)= sin(2\theta)[/itex]. The more general formula follows from the identites for sin(\theta+ \phi) and cos(\theta+ \phi).

Even simpler is to note that cos(\theta)+ i sin(\theta)= e^{i\theta} so that (cos(\theta)+ i sin(\theta)^n= (e^{i\theta})^n= e^{ni\theta}.