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Suppose that $0 \le \theta \lt 2\pi$. Determine the polar form of the following complex numbers: $$z = \sin\theta + i(1+ \cos\theta)$$ $$w = \cos\theta + \sin\theta +i(\sin\theta - \cos\theta)$$

The polar form of a complex number is given by $$x = |x|(\cos \theta +i\sin \theta)$$ I tried solving it with $(b/a) = \tan\theta$ but it yields an awkward result.

From $z$ I got $|z|=2\cos(\theta/2)$ which is not bad but when I tried to get the argument of $z$, looks like I’m missing some identity or something. $$\tan \phi = \frac{1 +\cos\theta}{sin\theta}$$ I did tried with the identity of the half angle tangent, but was not the same, I think.

Happens almost the same with $w$.

Note: $|w| =2 ^ {1/2}$

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Hint for $z$ \begin{eqnarray*} \cos(\theta)=2 \cos^2(\theta/2)-1 \\ \sin(\theta) = 2 \sin(\theta/2) \cos(\theta/2). \end{eqnarray*} Hint for $w$: Use the $\cos$ and $\sin$ addition formulae and \begin{eqnarray*} w &=& \cos \theta + \sin \theta +i(\sin \theta - \cos \theta) \\ &=&\sqrt{2} \left( \cos(\theta) \cos(\pi/4) + \sin(\theta) \sin(\pi/4) \\+i( \sin(\theta) \cos(\pi/4) - \cos(\theta) \sin(\pi/4)) \right). \end{eqnarray*}

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I have used $$1+\cos \theta =2\cos ^2 (\theta/2)$$ along with the usual angle addition formula to get $$z= \sin \theta +i(1+\cos \theta)$$

$$ |z|=2\cos(\theta/2)$$ and $$ arg (z) = ( \pi/2 -\theta /2)$$

Also for $w$ we have $$w=\cos (\theta) + \sin ( \theta ) +i(\sin (\theta) - \cos( \theta ))$$

$$|w| = \sqrt 2$$

$$ arg(w) = \theta -\pi/4$$

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  • $\begingroup$ Thank you for the answer @Mohammad. I’m sorry but according to the hint given by Donald Splutterwit, I got this $arg(z) = \cot^2 (\theta/2)$, may be is the reason you have that argument. Could you help me understand your result? $\endgroup$ – Octavio Berlanga Oct 2 at 15:51
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    $\begingroup$ @OctavioBerlanga let $ \alpha=arg(z)$ we have $\cos( \alpha) = \frac {\sin \theta }{2\cos (\theta /2)}= \sin(\theta /2)$ and $\sin \alpha = \frac {1+\cos \theta }{2\cos (\theta /2)} = \cos (\theta /2)$ hence $\alpha =( \pi/2-\theta /2)$ $\endgroup$ – Mohammad Riazi-Kermani Oct 2 at 16:01
  • $\begingroup$ Thank you @Mohammad $\endgroup$ – Octavio Berlanga Oct 2 at 16:59

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