Write the complex number $1 + \cos \alpha +i \sin \alpha$ in polar form

Question

Write the complex number $z=1 + \cos \alpha +i \sin \alpha$ in polar form

Here is what I got so far,

$r= |z|=((1+\cos \alpha)^2 +\sin^2 \alpha)^{1/2}$

this is not pretty number but I can simplify it a little bit, but what abour $arg(z)$

$arg(z)=arctan (\frac {\sin \alpha}{1+ \cos \alpha}$)

I'm not sure if this make sense to put into the polar form.

I also tried to break this in to 2 parts

$z=z_1 +z_2 $ where $z_1 =1=1(\cos 0 +i \sin 0)$ and $z_2= 1(\cos \alpha +i \sin \alpha)$ but that doesn't get me any where. Any help will be much appreciated.

Answer 1

We have $1+\cos\alpha+i(\sin\alpha)$

so, arg$(z)=\arctan\dfrac{\sin\alpha}{1+\cos\alpha}$

If $1+\cos\alpha=\sin\alpha=0\iff\alpha=(2n+1)\pi,$ where $n$ is any integer

arg$(z)$ will be undefined

Otherwise, arg$(z)=\arctan\dfrac{2\cos\dfrac\alpha2\sin\dfrac\alpha2}{2\cos^2\dfrac\alpha2}=\arctan\left[\tan\left(\dfrac\alpha2\right)\right]$

Now use the definition

and adjust $\dfrac\alpha2$ to $\beta=m\pi+\dfrac\alpha2$ for some integer $m$ so that $\beta$ lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$

Answer 2

i will use little bit geometry to get at the answer. let us keep $-\pi < \alpha < \pi$

let the point $o = 0, a = 1, u = \cos \alpha + i \sin \alpha$ and the point $z = 1 + u$ observe that $|u| = 1$ and the triangle $oaz$ is an isosceles triangle so the angle $aoz = \dfrac{\alpha}{2}$ and the side $|z| = 2\sin(\alpha/2)$
therefore in polar form $$ z = 2 \sin(\alpha/2) e^{i\alpha/2} \text{ if } 0 \le \alpha < \pi $$

$$ z = 2 \sin(-\alpha/2) e^{i\alpha/2} \text{ if } -\pi \le \alpha < 0 $$