Trig Identities when converting to polar form

Question

Been practicing conversions from rectangular form to polar form but got stuck on one question.

rectangular form: $\frac{1}{1+j\omega}$

polar form: $r = \left|\frac{1}{1+j\omega}\right|$, $\phi = -\tan^{-1}(\omega)$

Though I can't figure out how $\phi$ was found for this. The formula being $\phi = \tan^{-1}(\frac{y}{x})$. could someone show me the derivation for $\phi$?

Answer 1

Given the complex number $$z=\frac{1}{1+j\omega}$$

• To find the rectangular form, "rationalize" the denominator. That is $$z=\frac{1}{1+j\omega}\cdot \frac{1-j\omega}{1-j\omega}=\frac{1-j\omega}{1+\omega^2}=\frac{1}{1+\omega^2}-j\frac{\omega}{1+\omega^2}=x+jy$$ Thus \begin{align} x&=\frac{1}{1+\omega^2}\\ y&=-\frac{\omega}{1+\omega^2} \end{align}

• To find the polar form, find the modulus and the angle. \begin{align} r&=\left| z\right|=\left| \frac{1}{1+j\omega}\right|=\frac{\left| 1\right|}{\left| 1+j\omega\right|}=\frac{1}{\sqrt{1+\omega^2}}\\ \phi&=\angle{z}=\angle{\frac{1}{1+j\omega}}=\angle{1}-{\angle{1+j\omega}}=0-\tan^{-1}\left(\omega \right)=-\tan^{-1}\left(\omega \right) \end{align}

You can verify that $r=\sqrt{x^2+y^2}$ and $\phi=\tan^{-1}\left(\frac{y}{x} \right)$.