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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.

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  • $\begingroup$ You introduce confusion with two different $\theta$'s. $\endgroup$ – Yves Daoust Jun 25 at 8:15
  • $\begingroup$ My personal crusade: note that you have to be very careful when using $\tan^{-1}$ to find $\theta$ since $\tan^{-1}$ only gives a value in $\left( -\frac{\pi}{2},\frac{\pi}{2} \right)$. $\endgroup$ – Taladris Jun 25 at 9:00
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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?

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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?

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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*}

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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,

$$r=2|\sin\phi|.$$

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