Complex Conjugation & Division (Cambridge International A-Level Mathematics)

Complex conjugation & division

When dividing complex numbers, we can use the complex conjugate to make the denominator a real number, which makes carrying out the division much easier.

What is a complex conjugate?

• For a given complex number $z=a+bi$, the complex conjugate of$z$ is denoted as$z^{*}$, where $z^{*}=a-bi$

• If $z=a-bi$ then $z^{*}=a+bi$

• You will find that:

  • $z+z^{*}$ is always real because $(a+bi)+(a-bi)=2a$

    • For example: $(6+5i)+(6-5i)=6+6+5i-5i=12$

  • $z-z^{*}$ is always imaginary because $(a+b \mathrm{i} )− ( a−b \mathrm{i} ) =2b \mathrm{i}$

    • For example: $(6+5i)-(6-5i)=6-6+5i-(-5i)=10i$

  • $z\times z^{*}$ is always real because $(a+b \mathrm{i} )(a−b \mathrm{i} )=a²+a b \mathrm{i} −a b \mathrm{i} −b² \mathrm{i} ²=a²+b²$ (as $i^{2}=-1$)

    • For example: $(6+5i)(6-5i)=36+30i--30i-25i^{2}=36--25(-1)=61$

How do I divide complex numbers?

• When we divide complex numbers, we can express the calculation in the form of a fraction, and then start by multiplying the top and bottom by the conjugate of the denominator:

  • $\frac{a+bi}{c+di}=\frac{a+bi}{c+di}\times\frac{c-di}{c-di}$

• This ensures we are multiplying by 1; so not affecting the overall value

• This gives us a real number as the denominator because we have a complex number multiplied by its conjugate ($zz^{\ast}$)

• This process is very similar to “rationalising the denominator” with surds

Worked Example

Exam Hint

• We can speed up the process for finding $zz\ast$by using the basic pattern of (x+a)(x−a)=x²−a²
• We can apply this to complex numbers: $(a+b \mathrm{i} )(a−b \mathrm{i} )=a²−b² \mathrm{i} ²=a²+b²$ (using the fact that $i^{2}=-1$)
• So $3+4i$ multiplied by its conjugate would be $3^{2}+4^{2}=25$