0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ $z = r\cdot e^{i\varphi}$. $$\frac{z}{w} = \frac{r\cdot e^{i\varphi}}{\rho\cdot e^{i\psi}}.$$ Hmmmm. $\endgroup$ – Daniel Fischer Jul 31 '14 at 14:13
  • $\begingroup$ @DanielFischer Ummm...... Completely lost you there. $\endgroup$ – Gummy bears Jul 31 '14 at 14:49
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jul 31 '14 at 14:54
  • $\begingroup$ 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? $\endgroup$ – Gummy bears Aug 1 '14 at 16:06
  • $\begingroup$ Sure, $(r\cdot e^{i\varphi})(\rho \cdot e^{i\psi}) = (r\rho)\cdot e^{i\varphi}e^{i\psi}$. $\endgroup$ – Daniel Fischer Aug 1 '14 at 16:14
1
$\begingroup$

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 $\left(\frac{r_1}{r_2}\right) e^{i ( \theta_1 - \theta_2)}$.

$\endgroup$
1
$\begingroup$

$\def\cis{\operatorname{cis}}$ 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

$\endgroup$
  • 2
    $\begingroup$ To be clear, $cis(x) = e^{ix} = \cos(x) + i \sin(x)$ $\endgroup$ – LucasVB Jul 31 '14 at 14:21
  • $\begingroup$ @MJD - Thanks for the edit. I don't have the \cis command in Lyx but its on Math.SE (apparently) $\endgroup$ – Belgi Jul 31 '14 at 14:34
  • $\begingroup$ I had to define it specially for your post, using \def. $\endgroup$ – MJD Jul 31 '14 at 14:41
  • $\begingroup$ Waittt..... So I can basically subtract the angles? (Extending from this, I suppose in multiplication I can add the angles?) $\endgroup$ – Gummy bears Jul 31 '14 at 14:48
  • 1
    $\begingroup$ @Gummybears $e^{i \theta}= \cos{\theta} + i \sin{\theta}$. $\endgroup$ – snulty Aug 1 '14 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.