# Absolute value of complex number

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

• What are $\omega$, $L$, $C$, $G$? – martini Mar 15 '16 at 13:33
• Could you clarify what $\omega, L, C$ and $G$ are? – Claudius Mar 15 '16 at 13:33
• They are just constants from an eletrical circuit. – Vinícius Lopes Simões Mar 15 '16 at 13:36
• 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. – Antinous Mar 15 '16 at 13:36
• That is it! Thank you!! – Vinícius Lopes Simões Mar 15 '16 at 13:38

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 $j^2=i^2=-1$:
• $$|z|=\sqrt{\Re^2(z)+\Im^2(z)}$$
$$\left|\frac{1-\omega^2\text{LC}}{1-\omega^2\text{LC}+j\omega\text{LG}}\right|=\frac{\left|1-\omega^2\text{LC}\right|}{\left|1-\omega^2\text{LC}+j\omega\text{LG}\right|}=\frac{1-\omega^2\text{LC}}{\sqrt{\left(1-\omega^2\text{LC}\right)^2+\left(\omega\text{LG}\right)^2}}=$$ $$\frac{\sqrt{\left(1-\omega^2\text{LC}\right)^2}}{\sqrt{\left(1-\omega^2\text{LC}\right)^2+\left(\omega\text{LG}\right)^2}}=\sqrt{\frac{\left(1-\omega^2\text{LC}\right)^2}{\left(1-\omega^2\text{LC}\right)^2+\left(\omega\text{LG}\right)^2}}=\sqrt{\frac{1}{1+\left(\frac{\omega\text{LG}}{1-\omega^2\text{LC}}\right)^2}}=$$ $$\frac{1}{\sqrt{1+\left(\frac{\omega\text{LG}}{1-\omega^2\text{LC}}\right)^2}}$$
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.