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: $|w| =2 ^ {1/2}$
Best Answer
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*}