Absolute value of complex number

Question

This question might be very simple, but I can't visualize how to get the absolute value of this complex number ($j$ is the imaginary unit):

$$\frac{1-\omega^2LC}{1-\omega^2LC+j\omega LG}$$

Thanks

Answer 1

Assuming all other symbols are real numbers, it might help to first multiply top and bottom by the complex conjugate of the denominator, then expand the denominator. This will give you a complex number of the form $x+jy$, which you should then be able to find the modulus.

Answer 2

Notice, assuming $z\in\mathbb{C}$ and $\omega\space\wedge\space\text{L}\space\wedge\space\text{C}\space\wedge\space\text{G}\in\mathbb{R}$ and $j^2=i^2=-1$:

• $$|z|=\sqrt{\Re^2(z)+\Im^2(z)}$$

---

So, solving your question:

$$\left|\frac{1-\omega^2\text{LC}}{1-\omega^2\text{LC}+j\omega\text{LG}}\right|=\frac{\left|1-\omega^2\text{LC}\right|}{\left|1-\omega^2\text{LC}+j\omega\text{LG}\right|}=\frac{1-\omega^2\text{LC}}{\sqrt{\left(1-\omega^2\text{LC}\right)^2+\left(\omega\text{LG}\right)^2}}=$$ $$\frac{\sqrt{\left(1-\omega^2\text{LC}\right)^2}}{\sqrt{\left(1-\omega^2\text{LC}\right)^2+\left(\omega\text{LG}\right)^2}}=\sqrt{\frac{\left(1-\omega^2\text{LC}\right)^2}{\left(1-\omega^2\text{LC}\right)^2+\left(\omega\text{LG}\right)^2}}=\sqrt{\frac{1}{1+\left(\frac{\omega\text{LG}}{1-\omega^2\text{LC}}\right)^2}}=$$ $$\frac{1}{\sqrt{1+\left(\frac{\omega\text{LG}}{1-\omega^2\text{LC}}\right)^2}}$$