quotient of complex numbers?

Question

so I was wondering if you have two different equations having denominators $2+i$ and $2-i$ respectively how came the denominator of the quotient in standard form is $5$ for both equations? I tought the denominator was given by $a^2 + b^2$ for $z/w$? ? I mean looking at the first "denominator" wouldnt that be $3$? since $4 + i^2$ and $i^2 =-1$?

$$\begin{align} \ z &= x+yi \\ \ w &= a + bi \\ \end{align} $$

I was also wondering about something else and thats how to simplify $(2+i)^3$ ? I got something like $\sqrt{5} (\cos(3 \tan^{-1} (1/2) + i \sin(3 \tan^{-1} (1/2))$ which gave me $2/5 + (11/5)i$, but in the solution is says only $2 + 11i$??

You should be thinking of "$a+b\mathrm{i}$". $2 + \mathrm{i}$: $2^2+1^2 = 5$. $2 - \mathrm{i}$: $2^2+(-1)^2 = 5$. — Eric Towers, Jun 13 '14 at 20:24
ahh ok thanks, no wonder I was confused then :O, and also @user7530 how would you simplify (2+i)^100 then? cuz im pretty sure there is an easier way using the angle and modulus or whatever :P? — yyzzer1234, Jun 13 '14 at 20:26
Also notice that $\sqrt{5}\cos\left(3 \tan^{-1} \frac{1}{2}\right) = \frac{2}{5}$ so it's possible you already have the right answer. — user7530, Jun 13 '14 at 20:28
FK! I figured it out, it was just one simple mistake and thats cuz I forgot to raise root 5^3 :((( — yyzzer1234, Jun 13 '14 at 20:38
This is called rationalizing the denominator. See here for a general viewpoint and some interesting exmaples. — Bill Dubuque, Jun 13 '14 at 23:36
"Equation" is the wrong word here. "Expressions" is correct, but in this case I would have been more specific and said "fractions". — Michael Hardy, Jun 16 '14 at 01:16

beep-boop · Answer 1 · 2014-06-16T01:12:39.227

Here's a general trick to evaluate $\frac{a+bi}{c+di}$ in the form $x+yi$ (i.e. how to divide one complex number by another).

We use the fact that $(c+di)(c-di)\equiv c^2+d^2 $ (prove it yourself!)

$\color{green}{\frac{a+bi}{c+di}}=\frac{a+bi}{c+di}\cdot \underbrace{\left(\frac{c-di}{c-di}\right)}_{1}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{ac-adi+bci+bd}{c^2+d^2}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}=\color{green}{\frac{ac+bd}{c^2+d^2}+\left(\frac{bc-ad}{c^2+d^2}\right)i}$

score 2 · Accepted Answer · answered Jun 13 '14 at 20:28

The standard trick for simplifying quotients of complex number is to multiply both the numerator and the denominator by the conjugate of the denominator.

Let $w \neq 2+\mathrm{i}$ and $z\neq 2-\mathrm{i}$ be two complex numbers. Notice that $2+\mathrm{i}$ is the conjugate of $2-\mathrm{i}$ and vice-versa. Consider the following:

$$\frac{w}{2+\mathrm{i}} = \frac{w(2-\mathrm{i})}{(2+\mathrm{i})(2-\mathrm{i})}=\frac{w(2-\mathrm{i})}{4-2\mathrm{i}+2\mathrm{i}+1}=\frac{w(2-\mathrm{i})}{5}$$

$$\frac{z}{2-\mathrm{i}} = \frac{z(2+\mathrm{i})}{(2-\mathrm{i})(2+\mathrm{i})}=\frac{z(2+\mathrm{i})}{4+2\mathrm{i}-2\mathrm{i}+1}=\frac{z(2+\mathrm{i})}{5}$$

quotient of complex numbers?

2 Answers2