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New Zealand
Level 8 - NCEA Level 3

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$614i
$-15-3i$153i $-15+3i$15+3i
$1-i$1i $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×45i45i
  $=$= $\frac{\left(2+3i\right)\left(4-5i\right)}{\left(4+5i\right)\left(4-5i\right)}$(2+3i)(45i)(4+5i)(45i)
  $=$= $\frac{8+12i-10i-15i^2}{16+20i-20i-25i^2}$8+12i10i15i216+20i20i25i2
  $=$= $\frac{8+2i-15\left(-1\right)}{16-25\left(-1\right)}$8+2i15(1)1625(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×ii
  $=$= $\frac{-4i}{-i^2}$4ii2
  $=$= $\frac{-4i}{1}$4i1
  $=$= $-4i$4i


Example 2
$\frac{1-i}{1+i}$1i1+i $=$= $\frac{1-i}{1+i}\times\frac{1-i}{1-i}$1i1+i×1i1i
  $=$= $\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$(1i)(1i)(1+i)(1i)
  $=$= $\frac{1-2i+i^2}{1+1}$12i+i21+1
  $=$= $\frac{1-2i-1}{2}$12i12
  $=$= $\frac{-2i}{2}$2i2
  $=$= $-i$i


Example 3
$\frac{3}{6+i}+\frac{2}{3-2i}$36+i+232i $=$= $\frac{3}{6+i}\times\frac{6-i}{6-i}+\frac{2}{3-2i}\times\frac{3+2i}{3+2i}$36+i×6i6i+232i×3+2i3+2i
  $=$= $\frac{3\left(6-i\right)}{37}+\frac{2\left(3+2i\right)}{13}$3(6i)37+2(3+2i)13
  $=$= $\frac{18-3i}{37}+\frac{6+4i}{13}$183i37+6+4i13
  $=$= $\frac{456+109i}{481}$456+109i481



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


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


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


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



Manipulate complex numbers and present them graphically


Apply the algebra of complex numbers in solving problems

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