# Phase of a complex number

How can I calculate the phase of the following complex number?

$$\omega, R ,C$$ are some positive constants.

$$\frac{1-i\omega RC}{2+2i\omega RC}$$

In my book the answer is $$2\arctan{(\omega\cdot RC})$$

But I cannot see why. Here is my calculation:

$$\frac{1-i\omega RC}{2+2i\omega RC}=\frac{\left(1-i\omega RC\right)\left(1-i\omega RC\right)}{2\left(1+i\omega RC\right)\left(1-i\omega RC\right)}=\frac{1}{2}\frac{1-2i\omega RC-\left(\omega RC\right)^{2}}{\left(1+\left(\omega RC\right)^{2}\right)}=\frac{1-\left(\omega RC\right)^{2}}{2\left(1+\left(\omega RC\right)^{2}\right)}-i\frac{\omega RC}{\left(1+\left(\omega RC\right)^{2}\right)}$$

So the phase should be

$$\angle=\arctan\left(\frac{\frac{\omega RC}{1+\left(\omega RC\right)^{2}}}{\left(\frac{1-\left(\omega RC\right)^{2}}{2\left(1+\left(\omega RC\right)^{2}\right)}\right)}\right)=\arctan\left(\frac{2\omega RC}{1-\left(\omega RC\right)^{2}}\right)$$

The second way I tried and Im not sure why it is a wrong way, is to subtract the phase of the denominator from the phase on the numerator, so that $$\angle=\angle\left(1-i\omega RC\right)-\angle\left(2+2i\omega RC\right)=\arctan\left(\omega RC\right)-\arctan\left(\omega RC\right)$$

Which is obviously wrong.

Any help would be appreciated. Thank in advance.

Write $$z=1-i \omega RC$$ then your expression is: $$\frac{1}{2} \frac{z}{z^*}$$ where $$*$$ denotes the complex conjugate. Writing $$z=r e^{i\theta}$$ you have $$\frac{1}{2} \frac{z}{z^*}=\frac{1}{2} e^{2i\theta}$$.
Since the phase of $$z$$ is $$-\arctan(\omega RC)$$, the phase of your expression is : $$-2 \arctan(\omega RC)$$. To recover this result from yours, you can use the formula $$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy} \right)+k\pi$$. You made an error of sign in your last lign, otherwise you get the correct answer.
By the way, one should be careful with the use of $$\arctan$$ to recover the phase of $$x+iy$$ since the signs of $$x$$ and $$y$$ should be taken into account : $$\arctan$$ takes values in $$]-\pi/2,\pi/2[$$ while the phase is generally defined over $$]-\pi,\pi]$$.