Dividing Complex Numbers
Today, we're exploring how to divide complex numbers. Buckle up, because I'm going to guide you through a practical example to make this concept crystal clear.
Let's start with the basics. The quotient of two complex numbers (a,b) and (c,d) is given by the pair $$ \frac{(a,b)}{(c,d)} = ( \ \frac{ac+bd}{c^2+d^2} \ , \ \frac{bc-ad}{c^2+d^2} \ ) $$
To understand this better, let's jump into an example.
Imagine you have two complex numbers, z1 and z2:
$$ z_1 = (4,5) $$
$$ z_2 = (2,3) $$
Now, to divide these two complex numbers, z1/z2, we simply apply the formula above.
$$ \frac{z_1}{z_2} = ( \ \frac{ac+bd}{c^2+d^2} \ , \ \frac{bc-ad}{c^2+d^2} \ ) $$
Next, we're going to substitute a=4 and b=5 in the formula since our dividend z1=(a,b)=(4,5).
$$ \frac{z_1}{z_2} = ( \ \frac{4 \cdot c+5 \cdot d}{c^2+d^2} \ , \ \frac{5 \cdot c-4 \cdot d}{c^2+d^2} \ ) $$
Then, replace c=2 and d=3 as our divisor is z2=(c,d)=(2,3)
$$ \frac{z_1}{z_2} = ( \ \frac{4 \cdot 2+5 \cdot 3}{2^2+3^2} \ , \ \frac{5 \cdot 2-4 \cdot 3}{2^2+3^2} \ ) $$
Following this, our equation evolves
$$ \frac{z_1}{z_2} = ( \ \frac{8+15}{4+9} \ , \ \frac{10-12}{4+9} \ ) $$
Which simplifies to
$$ \frac{z_1}{z_2} = ( \ \frac{23}{13} \ , \ \frac{-2}{13} \ ) $$
Ultimately resulting in
$$ \frac{z_1}{z_2} = ( \ 1.77 \ , \ -0.15 \ ) $$
Voilà! The quotient of the two complex numbers is the complex number (1.77,-0.15).
Feeling skeptical? Feel free to verify the result using Geogebra, Matlab, or any other mathematical software you prefer.
Alternatively, if the formula for division seems a bit intimidating, don't fret! There's another approach to solve the division of complex numbers.
First, let's switch our complex numbers into algebraic form
$$ z_1 = (4,5) = 4+5i $$
$$ z_2 = (2,3) = 2+3i $$
To write a complex number (a,b) in algebraic form, simply add the real part to the imaginary part of the number. Keep in mind, the first element (a) in the pair (a,b) is the real part, while the second element (b) is the imaginary part. The imaginary part is always multiplied by the imaginary unit (i). $$ (a,b) = a+ b \cdot i $$
Now, let's tackle the division of these two complex numbers in algebraic form:
$$ \frac{z_1}{z_2} = \frac{4+5i}{2+3i} $$
A handy trick here is to multiply and divide by the complex conjugate of the denominator, in this case, 2-3i:
$$ \frac{z_1}{z_2} = \frac{4+5i}{2+3i} \cdot \frac{2-3i}{2-3i} $$
Wondering what a complex conjugate is? The conjugate of a complex number z=a+bi is another complex number z'=a-bi, having the same real part and the opposite value of the imaginary part. $$ z = a+bi $$ $$ z'=a-bi $$
Now, it's time to get our hands dirty with some good old algebra
$$ \frac{z_1}{z_2} = \frac{(4+5i) \cdot (2-3i)}{(2+3i) \cdot (2-3i) } $$
$$ \frac{z_1}{z_2} = \frac{8-12i+10i-15i^2}{4-6i+6i-9i^2} $$
$$ \frac{z_1}{z_2} = \frac{8-2i-15i^2}{4-9i^2} $$
The square of the imaginary unit i2=-1 is negative one.
$$ \frac{z_1}{z_2} = \frac{8-2i-15 \cdot (-1)}{4-9 \cdot (-1)} $$
$$ \frac{z_1}{z_2} = \frac{8-2i+15}{4+9} $$
$$ \frac{z_1}{z_2} = \frac{23-2i}{13} $$
$$ \frac{z_1}{z_2} = \frac{23}{13} - \frac{2}{13} i $$
After several steps of simplification, we reach the same result:
$$ \frac{z_1}{z_2} = \frac{23}{13} - \frac{2}{13} i = ( \frac{23}{13} , - \frac{2}{13} ) = ( \ 1.77 \ , \ -0.15 \ ) $$
There you have it, the division of complex numbers broken down step-by-step. If you enjoyed this deep dive into complex numbers with Nigiara, make sure to stay tuned for more enlightening lessons.
