# 2.07 Complex conjugates and moduli of complex numbers

Lesson

### 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+14​i $6-\frac{1}{4}i$6−14​i $-15-3i$−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 $a^2-b^2$a2b2 (for the distribution $\left(a+bi\right)\left(a-bi\right)$(a+bi)(abi))

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}$1i1+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+232i

 $\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}$23i2+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

Remember!

$r=\left|z\right|=\left|x+iy\right|=\sqrt{x^2+y^2}$r=|z|=|x+iy|=x2+y2

#### Worked example

##### Question 7

If $z=-2+2i$z=2+2i and $w=4-i$w=4i 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|=22.

 $|w|$|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) $|z-w|$|zw|

 $|z-w|$|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=1216i and $w=3-4i$w=34i, find:

1. $\left|z\right|$|z|

2. $\left|w\right|$|w|

3. $\left|z+w\right|$|z+w|

4. $\left|z\right|+\left|w\right|$|z|+|w|

### Outcomes

#### MGSE9-12.N.CN.3

Find the conjugate of a complex number; use the conjugate to find the absolute value (modulus) and quotient of complex numbers.