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


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*}


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 $$w=\cos (\theta) + \sin ( \theta ) +i(\sin (\theta) - \cos( \theta ))$$

$$|w| = \sqrt 2$$

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.