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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.

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Why would you expect the right expression to be real? –  DonAntonio May 26 '13 at 12:15
    
No of course, just $r,\theta$ should be real. –  aziiri May 26 '13 at 14:32
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2 Answers

up vote 2 down vote accepted

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 $u,v$:

enter image description here

$$u=\cos 2t+3\cos t+3 \text{ (blue) },\quad v=\sin 2t+3\sin t \text{ (red) },\quad 0\leq t\leq \pi.$$

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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

$$z^2+z+1=\frac{(e^{it}+1)^3-1}{e^{it}}=e^{2it}+3e^{it}+3=$$

$$=(\cos 2t+3\cos t+3)+(\sin 2t+3\sin t)i$$

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