2.07 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+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 $a^2-b^2$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

QUESTION 5

QUESTION 6

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

 

 

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

QUESTION 9