Complex numbers in polar form

If we have two complex numbers, in polar form, as the numerator and denominator of a fraction, and we are asked to write them as a single complex number, is there an easier way to deal with them rather than the usual procedure? (By usual procedure I mean first expanding them by writing the value of each term and then realizing the denominator, etc.)

Thank you.

• $z = r\cdot e^{i\varphi}$. $$\frac{z}{w} = \frac{r\cdot e^{i\varphi}}{\rho\cdot e^{i\psi}}.$$ Hmmmm. – Daniel Fischer Jul 31 '14 at 14:13
• @DanielFischer Ummm...... Completely lost you there. – Gummy bears Jul 31 '14 at 14:49
• It was meant to lead you to see that you can apply the functional equation of the exponential function for the argument(s), and that you can just divide the moduli. – Daniel Fischer Jul 31 '14 at 14:54
• I see. @DanielFischer So from there I can clearly see that I can multiply the moduli and just add the functional arguments if the two complex numbers are in multiplication? – Gummy bears Aug 1 '14 at 16:06
• Sure, $(r\cdot e^{i\varphi})(\rho \cdot e^{i\psi}) = (r\rho)\cdot e^{i\varphi}e^{i\psi}$. – Daniel Fischer Aug 1 '14 at 16:14

Do you mean you have $\frac{r_1 e^{i \theta_1}}{r_2 e^{i \theta_2}}$? In that case you can divide them as you might expect to yield $\left(\frac{r_1}{r_2}\right) e^{i ( \theta_1 - \theta_2)}$.

$\def\cis{\operatorname{cis}}$ If we have the fraction $$\frac{r_{1}\cis(\theta_{1})}{r_{2}\cis(\theta_{2})}$$

then in polar form this is the complex number $$\frac{r_{1}}{r_{2}}\cis(\theta_{1}-\theta_{2})$$

It follows from the fact that $|\cis(\alpha)|=1$ hence $|r\cis(\alpha)|=r$ and from the trigonometric identities

• To be clear, $cis(x) = e^{ix} = \cos(x) + i \sin(x)$ – LucasVB Jul 31 '14 at 14:21
• @MJD - Thanks for the edit. I don't have the \cis command in Lyx but its on Math.SE (apparently) – Belgi Jul 31 '14 at 14:34
• I had to define it specially for your post, using \def. – MJD Jul 31 '14 at 14:41
• Waittt..... So I can basically subtract the angles? (Extending from this, I suppose in multiplication I can add the angles?) – Gummy bears Jul 31 '14 at 14:48
• @Gummybears $e^{i \theta}= \cos{\theta} + i \sin{\theta}$. – snulty Aug 1 '14 at 12:49