I'm trying to solve the following fraction to find out what omega $\omega$ will leave me with only the real parts, assuming I know the values L, C, and R.

$z = \dfrac{ \dfrac{L}{C} + \jmath\omega RL}{R + \jmath(\omega L - \dfrac{\omega}{\omega C})}$

The normal way I would solve it is by multiplying the denominator and numerator by the complex conjugate of the denominator and solve for the $\omega$ that will make the complex part of the new numerator zero.

However, in my text book they solve for $\omega$ by

Making the ratio of the complex / real part of the numerator = complex /real part of the denominator. That is :

$\dfrac{\omega RL}{\dfrac{L}{C}} = \dfrac{\omega L - \dfrac{\omega}{\omega C}}{R}$

Then solving for $\omega$. Both methods giving me the same answer.

I was wondering if any body could help me understand why this method works or if there is a name for this technique? It seems to be a much shorter approach than the conjugate method. Thanks for any help.


1 Answer 1


If I may, let me just forget the specific problem of your post and let us consider a complex number $$z=\frac{a+i b}{c+i d}$$ As you said, let us multiply by the conjugate of the denominator to get $$z=\frac{(a+i b) (c-i d)}{(c+i d) (c-i d)}=\frac{(a c + b d)+i (b c-a d)}{c^2+d^2}$$ Since you want that the number be a real, it then implies that $$b c - a d =0 \implies \frac {b}{a}= \frac {d}{c} \text{ or }\frac {a}{b}= \frac {c}{d}$$ Now, replace $a,b,c,d$ by your numbers.

  • $\begingroup$ Suppose you instead want the number to be imaginary, is the condition instead, $ac+bd=0$, so $\frac{a}{b}=-\frac{d}{c}$? $\endgroup$
    – JDoeDoe
    Commented Nov 12, 2018 at 20:40

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