New Zealand
Level 8 - NCEA Level 3

# Complex Conjugates

Lesson

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 surds.  We underwent a process called rationalising the denominator, you can refresh here if you need to.

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 look at this question as an example. $\frac{2+3i}{4+5i}$2+3i4+5i

So, to tackle the division we

• multiply by a fraction that is equivalent to 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 expansion $\left(a+bi\right)\left(a-bi\right)$(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.

#### Here are some more examples

##### Example 1
 $\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

##### Example 2
 $\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

##### Example 3
 $\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​

Activity

Try this for yourself before you check out the solutions.

Verify which of the following hold true, check with $z=1+i$z=1+i and $w=2−2i$w=22i.

Then prove for $z=a+bi$z=a+bi and $w=c+di$w=c+di

1. $\overline{z-w}=\overline{z}-\overline{w}$zw=zw
2. $\overline{zw}=\overline{z}\times\overline{w}$zw=z×w
3. $\overline{z^n}=\left(\overline{z}\right)^n$zn=(z)n  use (b) to help here
4. $\overline{\left[\frac{z}{w}\right]}=\frac{\overline{z}}{\overline{w}}$[zw]=zw
5. What is $z+\overline{z}$z+z ? Generalise
6. What is $z-\overline{z}$zz ? Generalise

(go here for the solution)

#### More Worked Examples

##### QUESTION 1

Find the value of $\frac{4+6i}{1+i}$4+6i1+i.

##### QUESTION 2

Find the value of $\frac{4+7i}{2+i}$4+7i2+i.

##### QUESTION 3

Find the value of $\frac{2-3i}{2+3i}$23i2+3i.

### Outcomes

#### M8-9

Manipulate complex numbers and present them graphically

#### 91577

Apply the algebra of complex numbers in solving problems