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.


1 Answer 1


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 $]-\pi,\pi]$.


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.