# Complex numbers resolving cos and sin

I've got a problem with finding main argument of these complex number. How can i evaluate this two examples?

$$\sin \theta - i\cos \theta$$

$$\frac{(1-i\tan \theta)}{1+\tan \theta}$$

• main argument?? Commented Nov 13, 2016 at 17:47
• In my country it is called so, in other words the angle which is then placed in trigonometry form. Commented Nov 13, 2016 at 17:49
• My guess is that this is a reference to the Arg(z) function Commented Nov 13, 2016 at 21:04
• Yes, I'm looking for Arg(z) Commented Nov 13, 2016 at 22:00

Since $\arg(zw)=\arg(z)+\arg(w)$, \begin{align} \arg(\sin(\theta)-i\cos(\theta)) &=\arg(-i)+\arg(\cos(\theta)+i\sin(\theta))\\ &=\theta-\frac\pi2 \end{align} and \begin{align} \arg\left(\frac{1-i\tan(\theta)}{1+\tan{\theta}}\right) &=\arg(\cos(-\theta)+i\sin(-\theta))-\arg(\sin(\theta)+\cos(\theta))\\ &=-\theta+\pi[\sin(\theta)+\cos(\theta)\lt0] \end{align} where $[\dots]$ are Iverson Brackets. Note that $\arg(z)$ is determined mod $2\pi$.
It's not so clear what you mean, perhaps it's meant (where $z$ is a complex variable, here $z:=r(\cos\phi +i\sin\phi)$ with $r\in\mathbb{R}$):
$\displaystyle \sin\phi-i\cos\phi=-i\frac{z}{|z|}$
$\displaystyle \frac{1-i\tan\phi}{1+\tan\phi}=\frac{2\overline{z}}{(1-i)z+(1+i)\overline{z}}$