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 $(cos(\theta)+ i sin(\theta)^n= (e^{i\theta})^n= e^{ni\theta}$.