trigonometric representation of a complex number. Let $z=e^{it}+1$ where $0\leq t\leq \pi$, Find the trigonometric representation of $z^2+z+1$.
(The trigonometric representation should be in the form of : $r(\cos \theta +i \sin \theta)$, where $r,\theta \in \mathbb{R}$ and $r>0$.

What I have done : $$z^2+z+1=e^{2it}+3z=(\cos (t)+i \sin (t)) (-2 i \sin (t)+4 \cos (t)+3).$$
The problem is that the expression in the right bracket is not real always.
 A: The complex number $a+bi$ can be written in trigonometric form as
$$a+bi=\sqrt{a^{2}+b^{2}}(\cos \theta +i\sin \theta ),\qquad\text{with } \tan \theta =\frac{b}{a}.$$
For $z=e^{it}+1=(\cos t+1)+i\sin t$  we have 
\begin{eqnarray*}
z^{2}+z+1 &=&\left( e^{it}+1\right) ^{2}+(e^{it}+1)+1= (e^{i2t}+2e^{it}+1)+(e^{it}+1)+1 \\
&=&e^{2it}+3e^{it}+3 \\
&=&((\cos 2t+1)+i\sin 2t)+3\left( (\cos t+1)+i\sin t\right) +3 \\
&=&(\cos 2t+3\cos t+3)+i\left( \sin 2t+3\sin t\right)  \\
&=&u+iv,\qquad u=\cos 2t+3\cos t+3,v=\sin 2t+3\sin t \\
&=&r\left( \cos \theta +i\sin \theta \right) ,
\end{eqnarray*}
where
\begin{eqnarray*}
r &=&\sqrt{u^{2}+v^{2}}=\sqrt{(\cos 2t+3\cos t+3)^{2}+\left( \sin 2t+3\sin
t\right) ^{2}} \\
\tan \theta  &=&\frac{v}{u}.
\end{eqnarray*}
For $0\leq t\leq \pi ,v=\sin 2t+3\sin t\geq 0$ and $u=\cos 2t+3\cos t+3\geq 0
$. So
$$
\begin{equation*}
\theta =\arctan  \frac{v}{u} =\arctan \left( \frac{\sin
2t+3\sin t}{\cos 2t+3\cos t+3}\right) \geq 0,
\end{equation*}
$$
because $w=z^2+z+1=u+iv=\operatorname{Re}(w)+i\operatorname{Im}(w)$ is in the first quadrant.  
Plots of $u,v$:

$$u=\cos 2t+3\cos t+3 \text{ (blue) },\quad v=\sin 2t+3\sin t \text{ (red) },\quad 0\leq t\leq \pi.$$
A: Another approach, perhaps a little more elementary:
$$a^3-b^3=(a-b)(a^2+ab+b^2)\implies a^2+ab+b^2=\frac{a^3-b^3}{a-b}$$
Well, now just put $\,a=z\;,\;b=1\,$ above , and get
$$z^2+z+1=\frac{z^3-1}{z-1}$$
and since $\,z=e^{it}+1\,$ , we get
$$z^2+z+1=\frac{(e^{it}+1)^3-1}{e^{it}}=e^{2it}+3e^{it}+3=$$
$$=(\cos 2t+3\cos t+3)+(\sin 2t+3\sin t)i$$
