Find the angle of complex number Let
$$
z = \frac{a-jw}{a+jw}.
$$
Then the angle of $z$ is
$$
-\tan^{-1}z\left(\frac{w}{a}\right) -\tan^{-1}\left(\frac{w}{a}\right) = -2\tan^{-1}\left(\frac{w}{a}\right).
$$
How is that so?
 A: They are conjugate complex numbers
$$ \dfrac{r \cdot e ^{-i \theta }}{{r \cdot e ^{i \theta }}{} } =e ^{- 2 i \theta } $$  
A: Multiply numerator and denomination by the conjugate of  a+jw  i.e.  a-jw
So the numerator becomes  a2 -w2 - 2awj and the denominator will be real and you'll be able to solve it and get the final argument.
OR
Taking numerator as z1 and denominator as z2 
We know  z1=|z1|eiarg(z1)

Similarly z2=|z2|eiarg(z1)
Dividing we have, z=(|z1|/|z2|)ei(arg(z1)-arg(z2))
=>arg(z)= arg(z1)-arg(z2) 
Also, arg of a complex number is tan-1(complex part/real part) which in case of numerator will be tan-1(-w/a) = -tan-1(w/a) 
and for denominator , argument will be tan-1(w/a)
and since we are dividing than be another complex number which is denominator, we subtract their arguments to get the final argument.
A: I suppose that it could be easier to see writing $$z = \frac{a-jw}{a+jw}= \frac{a-jw}{a+jw} \times \frac{a-jw}{a-jw}=\frac{(a-jw)^2}{a^2+w^2}=\frac{a^2-w^2}{a^2+w^2}-\frac{2  a w}{a^2+w^2}j$$ which makes the angle to be such that $$\tan(\theta)=-\frac{2aw}{a^2-w^2}=-\frac{2\frac aw}{1-(\frac aw)^2}$$ and to remember the development of $\tan(2x)$.
A: Assuming $a,w\in\mathbb{R}$:
$$\arg\left(\frac{a-wi}{a+wi}\right)=$$
$$\arg\left(\frac{a-wi}{a+wi}\cdot\frac{a-wi}{a-wi}\right)=$$
$$\arg\left(\frac{(a-wi)^2}{a^2+w^2}\right)=$$
$$\arg\left(\frac{a^2-w^2-2awi}{a^2+w^2}\right)=$$
$$\arg\left(a^2-w^2-2awi\right)-\arg\left(a^2+w^2\right)=$$
$$\arg\left(a^2-w^2-2awi\right)-0=$$
$$\arg\left(a^2-w^2-2awi\right)=$$
$$\arg\left(a^2-w^2-2awi\right)=$$
$$\arg\left((a-wi)^2\right)=$$
$$\arg\left((a-wi)(a-wi)\right)=$$
$$\arg\left(a-wi\right)+\arg\left(a-wi\right)=$$
$$2\arg\left(a-wi\right)=$$
$$-2\tan^{-1}\left(\frac{w}{a}\right)$$
