If I have a complex fraction $\dfrac{a+bi}{c+di}$ and I want the magnitude, then will it be $\left|\dfrac{a+bi}{c+di}\right|=\dfrac{|a+bi|}{|c+di|}$?

Scratch that ... I just found the answer on another page; however, I'm still unclear why it's true?

  • $\begingroup$ @Ball Hint: use the properties of complex numbers division en.wikipedia.org/wiki/… . Compute both values and compare them. $\endgroup$ – yaa09d Apr 24 '14 at 2:56
  • 4
    $\begingroup$ A complex number can be represented as $re^{i\theta}$ where $r$ is magnitude, try using it on both ... numerator and denominator. $\endgroup$ – Santosh Linkha Apr 24 '14 at 3:12

You can make use of complex exponents. $$\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}=\frac{\rho_1e^{\mathrm{i} \varphi_1}}{\rho_2e^{\mathrm{i} \varphi_2}}=\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)}$$ where $\rho_1=\sqrt{a^2+b^2}, \rho_2=\sqrt{c^2+d^2}$ are the magnitudes and $\varphi_1=\arg\{a+\mathrm{i} \ b\},\varphi_2=\arg\{c+\mathrm{i} \ d\}$ are phases of $a+\mathrm{i} \ b$ and $c+\mathrm{i} \ d$ respectively.
Then since $\rho_1, \rho_2$ are real (and positive) and the absolute value of complex exponent is $1$: $$\left| \dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right|=\left|\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)} \right|=\left|\frac{\rho_1}{\rho_2}\right|\left|e^{\mathrm{i}(\varphi_1-\varphi_2)} \right|=\left|\frac{\rho_1}{\rho_2}\right|=\frac{\left|\rho_1\right|}{\left|\rho_2\right|}=\frac{\left|a+\mathrm{i} \ b\right|}{\left|c+\mathrm{i} \ d\right|}.$$ Moreover, using complex exponents it is easy to show that $$\arg\left\{\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right\}=\arg\left\{a+\mathrm{i} \ b\right\}-\arg\left\{c+\mathrm{i} \ d\right\}.$$ That is true, since $\arg\left\{\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right\}=\arg\left\{\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)}\right\}=\varphi_1-\varphi_2$.

  • $\begingroup$ Now that's a very concise and useful explanation, I'd totally forgotten about the exponential form for complex numbers. $\endgroup$ – Daniel B. Apr 24 '14 at 4:45

A simpler approach:

Let $z_1=a+bi$ and $z_2=c+di$. Since by properties of absolute value we have $|z_1z_2|=|z_1||z_2|,$ and the fact that $z_2(\frac{z_1}{z_2})=z_1$ then we have that $$\left|z_2\frac{z_1}{z_2}\right|=|z_1|\implies|z_2|\bigg|\frac{z_1}{z_2}\bigg|=|z_1|\implies \bigg|\frac{z_1}{z_2}\bigg|=\frac{|z_1|}{|z_2|}$$

  • $\begingroup$ Do the rules for an absolute value apply for a magnitude in all cases? |a+bi| = sqrt(a^2 + b^2) ... $\endgroup$ – Daniel B. Apr 24 '14 at 4:43
  • $\begingroup$ Yes, for every complex number it applies. $\endgroup$ – homegrown Apr 24 '14 at 4:48
  • $\begingroup$ This is actually how this property is proved in Ahlfor's Complex Analysis book as well. $\endgroup$ – homegrown Apr 24 '14 at 4:50
  • $\begingroup$ @DanielBall We always have $|z_1||z_2|=|z_1z_2|$; it's a simple computation, though maybe a bit tedious, and essential for a lot of stuff! $\endgroup$ – user98602 Apr 24 '14 at 4:50
  • 1
    $\begingroup$ Yes, absolute value is a specific case of a norm (magnitude). It is called the Euclidean norm. Since any complex number can be thought of as a vector in $\mathbb R^2$, then really what you have is for $z=a+bi$, $|z|=\sqrt{a^2+b^2}.$ This is because $z\bar z=|z|^2$, by definition. $\endgroup$ – homegrown Apr 24 '14 at 5:23

$\frac{a+bi}{c+di} = \frac{a+bi}{c+di} * \frac{c-di}{c-di} = i (\frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2})+\frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$. At this point, you should be able to get the magnitude easily. Yes, it'll be cumbersome computation wise, but that should be it.

Suppose $e = \frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2}$ and $f = \frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$

Then, $\|f + ei\| = \sqrt{f^2+e^2} = \sqrt{\frac{(bc-ad)^2}{(c^2+d^2)^2} + \frac{(ac+bd)^2}{(c^2+d^2)^2}} = \sqrt{\frac{2(a^2d^2+b^2c^2)}{(c^2+d^2)^2}}$ and you could take it from there.


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.