Finding the angle of a complex fraction

Question

If I have a complex fraction, say $$z=\frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$\arctan(\frac{b}{a})-\arctan(\frac{d}{c})?$$

For example, my textbook says that for the fraction $$z=\frac{j\omega}{1+j\omega},$$ the phase angle should be $$\arctan(\frac{1}{\omega}).$$ I'm not really understanding what method/identities they're using to arrive at this answer.Is there a standard way to find the phase angle of a complex fraction?

Answer 1

If $z = a+bj$ where $a$ and $b$ are real, then you can determine the phase angle as $\arctan\left(\frac{b}{a}\right)$ (as long as you are careful about the signs and quadrants), since $a+bj$ represents a point in the complex plane, and the phase angle is the angle this vector makes with the $x$-axis. So for the example from your textbook, \begin{align*} \frac{j\omega}{1+j\omega} = \frac{j\omega(1-j\omega)}{(1+j\omega)(1-j\omega)} = \frac{\omega^2 + j\omega}{1+\omega^2}. \end{align*} Then the phase angle is $\arctan\left(\frac{\omega}{\omega^2}\right) = \arctan\left(\frac{1}{\omega}\right)$, as the book says.

If you take the fraction you are given, $\frac{a+bi}{c+di}$ and rationalize the denominator, you should be able to write the phase angle as an arctangent just as above.

Answer 2

The argument of the complex number $\frac{a+bi}{c+di}$ is just the difference between the arguments of $a+bi$ and $c+di$.

So you really just need to know how to compute the argument of any complex number $a+bi$ (and then you simply compute a difference of two arguments).

• If $a > 0$, this is indeed $\arctan \left(\frac{b}{a}\right)$.
• If $a < 0$, this is $\arctan \left(\frac{b}{a}\right) \pm\pi$ depending on the sign of $b$ (I suggest you draw it to see).
• If $a = 0$, this is $\pm \frac{\pi}{2}$ depending depending on the sign of $b$.

Answer 3

If I have a complex fraction, say $$z=\frac{a+bi}{c+di}$$, how can I find the phase angle of this? Could I just calculate it using $$\arctan(\frac{b}{a})-\arctan(\frac{d}{c})?$$

Yes, you can. In fact, I can show you that find phasor angle of the rationalized $z$ is the same as the phasor angle of the numerator minus that of the denominator.

First the rationalization approach:

$$\angle \frac{a+bi}{c+di} = \angle \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \angle\frac{(ac+bd)+(bc-ad)i}{c^2+d^2} = arctan(\frac{bc-ad}{ac+bd})$$

Now calculate the numerator and denominator separately and use Sum of Arctangents: $$\arctan a + \arctan b = \arctan \dfrac {a + b} {1 - a b}$$

$$\angle(a+bi)-\angle(c+di) = arctan(\frac{b}{a})-arctan(\frac{d}{c}) = arctan(\frac{\frac{b}{a}-\frac{d}{c}}{1+\frac{bd}{ac}})=arctan(\frac{bc-ad}{ac+bd})$$