12.02 Complex conjugates and moduli of complex numbers
Complex conjugates
A complex conjugate is defined as a number with equal magnitudes of real and imaginary parts, but opposite in sign.
Let's have a look at some conjugate pairings
| $6+\frac{1}{4}i$6+14i | $6-\frac{1}{4}i$6−14i |
| $-15-3i$−15−3i | $-15+3i$−15+3i |
| $1-i$1−i | $1+i$1+i |
| $26i$26i | $-26i$−26i |
| $7$7 | $7$7 |
There is a special symbol we use to denote the conjugate of $z$z. It is called zed bar, and looks like this 
.
We have dealt with a similar concept before when we were studying radicals. We underwent a process called rationalizing the denominator.
Just like this process, we use complex conjugates to help us define a process for division of complex numbers.
Instead of a formal division process, we use conjugates to turn the operation of division into an operation of multiplication.
Let's rationalize the denominator of the fraction: $\frac{2+3i}{4+5i}$2+3i4+5i
So, to tackle the division we
- multiply by a fraction that is equivalent to $1$1
- use the conjugate of the denominator to construct this fraction
- multiply the binomials on the numerator and denominators and simplify
- the denominator will always simplify to $a^2-b^2$a2−b2 (for the distribution $\left(a+bi\right)\left(a-bi\right)$(a+bi)(a−bi))
Let's finish off this question $\frac{2+3i}{4+5i}$2+3i4+5i
| $\frac{2+3i}{4+5i}$2+3i4+5i | $=$= | $\frac{2+3i}{4+5i}\times\frac{4-5i}{4-5i}$2+3i4+5i×4−5i4−5i |
| $=$= | $\frac{\left(2+3i\right)\left(4-5i\right)}{\left(4+5i\right)\left(4-5i\right)}$(2+3i)(4−5i)(4+5i)(4−5i) | |
| $=$= | $\frac{8+12i-10i-15i^2}{16+20i-20i-25i^2}$8+12i−10i−15i216+20i−20i−25i2 | |
| $=$= | $\frac{8+2i-15\left(-1\right)}{16-25\left(-1\right)}$8+2i−15(−1)16−25(−1) | |
| $=$= | $\frac{8+2i+15}{16+25}$8+2i+1516+25 | |
| $=$= | $\frac{23+2i}{41}$23+2i41 | |
| $=$= | $\frac{23}{41}+\frac{2i}{41}$2341+2i41 |
This last step may or may not be necessary. Sometimes it's easier to do further work when the real and imaginary parts are separate.
Worked examples
Question 1
Rationalize: $\frac{4}{i}$4i
| $\frac{4}{i}$4i | $=$= | $\frac{4}{i}\times\frac{-i}{-i}$4i×−i−i |
| $=$= | $\frac{-4i}{-i^2}$−4i−i2 | |
| $=$= | $\frac{-4i}{1}$−4i1 | |
| $=$= | $-4i$−4i |
Question 2
Rationalize: $\frac{1-i}{1+i}$1−i1+i
| $\frac{1-i}{1+i}$1−i1+i | $=$= | $\frac{1-i}{1+i}\times\frac{1-i}{1-i}$1−i1+i×1−i1−i |
| $=$= | $\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$(1−i)(1−i)(1+i)(1−i) | |
| $=$= | $\frac{1-2i+i^2}{1+1}$1−2i+i21+1 | |
| $=$= | $\frac{1-2i-1}{2}$1−2i−12 | |
| $=$= | $\frac{-2i}{2}$−2i2 | |
| $=$= | $-i$−i |
Question 3
Simplify: $\frac{3}{6+i}+\frac{2}{3-2i}$36+i+23−2i
| $\frac{3}{6+i}+\frac{2}{3-2i}$36+i+23−2i | $=$= | $\frac{3}{6+i}\times\frac{6-i}{6-i}+\frac{2}{3-2i}\times\frac{3+2i}{3+2i}$36+i×6−i6−i+23−2i×3+2i3+2i |
| $=$= | $\frac{3\left(6-i\right)}{37}+\frac{2\left(3+2i\right)}{13}$3(6−i)37+2(3+2i)13 | |
| $=$= | $\frac{18-3i}{37}+\frac{6+4i}{13}$18−3i37+6+4i13 | |
| $=$= | $\frac{456+109i}{481}$456+109i481 |
Practice questions
QUESTION 4
Find the value of $\frac{4+6i}{1+i}$4+6i1+i.
QUESTION 5
Find the value of $\frac{4+7i}{2+i}$4+7i2+i.
QUESTION 6
Find the value of $\frac{2-3i}{2+3i}$2−3i2+3i.
Complex moduli
The modulus of a complex number $x+yi$x+yi is the distance of the vector created from the origin to the point $\left(x,y\right)$(x,y). In other words, it is the size of the number.
The modulus is denoted using many different notations in various texts, websites, countries and schools. Here are a list of various notations.
- the letter $r$r
- $modz$modz
- $|z|$|z|
- $|x+iy|$|x+iy|
In this diagram, the point $P$P has been plotted. It corresponds to the complex number $3+4i$3+4i.
We can see the horizontal distance is $3$3, (the $x$x value) and the vertical distance is $4$4, (the $y$y value). The modulus is the length of the hypotenuse, (which is the length of the vector $P$P)
This distance can be found using the Pythagorean theorem.
| $r^2$r2 | $=$= | $x^2+y^2$x2+y2 |
| $r$r | $=$= | $\sqrt{x^2+y^2}$√x2+y2 |
| $=$= | $\sqrt{3^2+4^2}$√32+42 | |
| $=$= | $5$5 |
$r=\left|z\right|=\left|x+iy\right|=\sqrt{x^2+y^2}$r=|z|=|x+iy|=√x2+y2
Worked example
Question 7
If $z=-2+2i$z=−2+2i and $w=4-i$w=4−i then find
a) $|z|$|z|
| $|z|$|z| | $=$= | $|-2+2i|$|−2+2i| |
| $=$= | $\sqrt{(-2^2)+2^2}$√(−22)+22 | |
| $=$= | $\sqrt{8}$√8 | |
| $=$= | $2\sqrt{2}$2√2 |
b) $|z|^2$|z|2
| $|z|^2$|z|2 | $=$= | $|-2+2i|^2$|−2+2i|2 |
| $=$= | $\left(\sqrt{8}\right)^2$(√8)2 | |
| $=$= | $8$8 |
c) $|z^2|$|z2|
| $|z^2|$|z2| | $=$= | $\left|\left(-2+2i\right)^2\right|$|(−2+2i)2| |
| $=$= | $\left|\left(-2+2i\right)\left(-2+2i\right)\right|$|(−2+2i)(−2+2i)| | |
| $=$= | $\left|-4-4i-4i+4i^2\right|$|−4−4i−4i+4i2| | |
| $=$= | $\left|-8-8i\right|$|−8−8i| | |
| $=$= | $\sqrt{-8^2+-8^2}$√−82+−82 | |
| $=$= | $\sqrt{64+64}$√64+64 | |
| $=$= | $\sqrt{128}$√128 | |
| $=$= | $8\sqrt{2}$8√2 |
d) $|z|-|w|$|z|−|w|
From part (a), we know that $|z|=2\sqrt{2}$|z|=2√2.
| $|w|$|w| | $=$= | $\sqrt{4^2+-1^2}$√42+−12 |
| $=$= | $\sqrt{16+1}$√16+1 | |
| $=$= | $\sqrt{17}$√17 | |
| Therefore: | ||
| $|z|-|w|$|z|−|w| | $=$= | $2\sqrt{2}-\sqrt{17}$2√2−√17 |
which is approximately $1.29$1.29.
e) $|z-w|$|z−w|
| $|z-w|$|z−w| | $=$= | $\left|\left(-2-2i\right)-\left(4-i\right)\right|$|(−2−2i)−(4−i)| |
| $=$= | $\left|-2-4-2i+i\right|$|−2−4−2i+i| | |
| $=$= | $\left|-6-i\right|$|−6−i| | |
| $=$= | $\sqrt{(-6)^2+(-1)^2}$√(−6)2+(−1)2 | |
| $=$= | $\sqrt{37}$√37 |
Practice questions
QUESTION 8
Find the modulus of $-8+3i$−8+3i.
QUESTION 9
If $z=12-16i$z=12−16i and $w=3-4i$w=3−4i, find:
$\left|z\right|$|z|
$\left|w\right|$|w|
$\left|z+w\right|$|z+w|
$\left|z\right|+\left|w\right|$|z|+|w|