Find the polar representation of $z = \sin\theta+i(1+\cos\theta)$ 
Find the polar representation of the complex number $$z = \sin\theta+i(1+\cos\theta)$$ where $θ \in[0,2\pi)$.

I know how to write a formula in polar representation using hard numbers, e.g. $$ 
z = 3(\cos\pi+i\sin\pi)$$ and know that I need to find both my radius using the $\sqrt{a^2+b^2}$ formula, and to also find $\theta$ using $\tan^{-1}$.
The problem I have with this question is I am not sure how to do it without actual numbers. Need lots of help.
 A: Note that\begin{align}z&=\sin(\theta)+i\bigl(1+\cos(\theta)\bigr)\\&=2\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)+i\left(\cos^2\left(\frac\theta2\right)+\sin^2\left(\frac\theta2\right)+\cos^2\left(\frac\theta2\right)-\sin^2\left(\frac\theta2\right)\right)\\&=2\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)+2i\cos^2\left(\frac\theta2\right)\\&=2\cos\left(\frac\theta2\right)\left(\sin\left(\frac\theta2\right)+i\cos\left(\frac\theta2\right)\right)\\&=2\cos\left(\frac\theta2\right)\left(\cos\left(\frac\pi2-\frac\theta2\right)+i\sin\left(\frac\pi2-\frac\theta2\right)\right).\end{align}Can you take it from here?
A: Using your technique,
\begin{align}
r &= \sqrt{\sin^2 \theta + (1+\cos\theta)^2}\\
\phi &= \arctan\left(\frac{1+\cos\theta}{\sin\theta}\right)= \arctan\left(\cot (\theta/2)\right)
\end{align}
where $z = re^{i\phi}$. Can you simplify these expressions?
A: We have 
\begin{eqnarray*}
x= r \cos \phi =\sin \theta \\ 
y= r \sin \phi =1+ \cos \theta \\ 
\end{eqnarray*}
Now eliminate $\theta $ (using $\cos^2 \theta +\sin^2 \theta=1$)
\begin{eqnarray*}
( r \cos \phi )+(r \sin \phi -1)^2=1 \\ 
\end{eqnarray*}
and this gives 
\begin{eqnarray*}
r =2\sin \phi. \\ 
\end{eqnarray*}
A: You have
$$a=\sin\theta,\\b=1+\cos\theta$$ so you can apply the formulas you know.
$$r=\sqrt{\sin^2\theta+(1+\cos\theta)^2},\\\phi=\arctan\frac{1+\cos\theta}{\sin\theta}.$$
A more friendly form is obtained by a trigonometric transformation, giving after simplification
$$r=2\left|\cos\frac\theta2\right|,\\\phi=\arctan\cot\frac\theta2=\frac{\pi-\theta}2.$$
Finally,
$$r=2|\sin\phi|.$$
