0
$\begingroup$

I am trying to find the argument of $z=(sin(\theta) + i(1-cos(\theta))^2$ for $0 < \theta< \pi/2$ in its simplest form.

I've tried expanding it out:

\begin{align*} z&=(\sin^2 \theta + 2 i \sin \theta \cos \theta + i^2 (1-\cos \theta)^2)\\ &=-\cos(2 \theta) +2 \cos \theta - 2i \sin \theta - i \sin(2 \theta) - 1 \\ &=-\operatorname{cis}(2 \theta) + 2 \operatorname{cis}(\theta) -1 \end{align*}

so $z$ is a quadratic in a complex number, im not sure if this helps in finding arg(z) though? I'm a bit rusty with my complex numbers

$\endgroup$
  • $\begingroup$ Hint: Double angle formulas. $\endgroup$ – Did Oct 17 '15 at 12:57
1
$\begingroup$

HINT:

$$\sin\theta+i(1-\cos\theta)=2\sin\dfrac\theta2\left(\cos\dfrac\theta2+i\sin\dfrac\theta2\right)$$

Now use de Moivre's formula

$\endgroup$
1
$\begingroup$

Asumming $0<x<\frac{\pi}{2}$:

$$\arg\left((\sin(x) + i(1-\cos(x))^2\right)=$$ $$\arg\left(\sin^2\left(\frac{x}{2}\right)\left(4\cos(x)+4i\sin(x)\right)\right)=$$ $$\arg\left(\left(2\cos(x)-\cos(2x)-1\right)+\left(2\sin(x)-\sin(2x)\right)i\right)=$$ $$\tan^{-1}\left(\frac{2\sin(x)-\sin(2x)}{2\cos(x)-\cos(2x)-1}\right)=$$ $$\tan^{-1}\left(\tan(x)\right)$$

$\endgroup$

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.