10
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ @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. $\endgroup$ Apr 24 '14 at 3:12
13
$\begingroup$

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 $\arg\left\{\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right\}=\arg\left\{\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)}\right\}=\varphi_1-\varphi_2$.

$\endgroup$
1
  • $\begingroup$ 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
10
$\begingroup$

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 $$\left|z_2\frac{z_1}{z_2}\right|=|z_1|\implies|z_2|\bigg|\frac{z_1}{z_2}\bigg|=|z_1|\implies \bigg|\frac{z_1}{z_2}\bigg|=\frac{|z_1|}{|z_2|}$$

$\endgroup$
7
  • $\begingroup$ 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
3
$\begingroup$

$\frac{a+bi}{c+di} = \frac{a+bi}{c+di} * \frac{c-di}{c-di} = i (\frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2})+\frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$. 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.

$\endgroup$

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