International ASFurther MathsEdexcelFurther Pure 1Revision NotesComplex NumbersOperations with Complex NumbersEquating Real & Imaginary Parts

Equating Real & Imaginary Parts (Edexcel International AS Further Maths)

Revision Note

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Equating Real & Imaginary Parts

How do I equate real and imaginary parts?

If two complex expressions are equal, then
- their real parts are equal
- and their imaginary parts are equal
If $a + b i = 8 - 10 i$ where $a$ and $b$ are real
- then $a = 8$ (equating real parts)
- and $b = - 10$ (equating imaginary parts)

How do I solve equations using real and imaginary parts?

Introduce $z = x + y i$ for expressions in $z$
For example, solve $z + 2 z * = 12 - i$
- Let $z = x + y i$
- Substitute in
  - $(x + y i) + 2 (x - y i) = 12 - i$
- Expand and collect terms
  - $3 x - y i = 12 - i$
- Equate real and imaginary parts
  - $3 x = 12$ gives $x = 4$
  - $- y = - 1$ gives $y = 1$
- Substitute back into $z$
  - $z = 4 + i$

How do I use real and imaginary parts with modulus signs?

If $z = x + y i$ then $| z | = \sqrt{x^{2} + y^{2}}$
Helpful modulus rules are:
- $| z_{1} z_{2} | = | z_{1} | | z_{2} |$
- $|\frac{z_{1}}{z_{2}}| = \frac{|z_{1}|}{|z_{2}|}$
- $| z | = | z * |$
- $z z * = z * z = | z |^{2}$
  - Proved using $z = x + y i$ and $z * = x - y i$
For example, find $p$ if $z_{1} = p + p i$ , $z_{2} = 2 i$ and $| z_{1} z_{2} | = 6 \sqrt{2}$
- Use $| z_{1} z_{2} | = | z_{1} | | z_{2} |$
  - $| p + p i | | 2 i | = 6 \sqrt{2}$
- Use $| z | = \sqrt{x^{2} + y^{2}}$
  - $\sqrt{p^{2} + p^{2}} \times 2 = 6 \sqrt{2}$ so $2 \sqrt{2 p^{2}} = 6 \sqrt{2}$
- Square both sides and simplify
  - $p^{2} = 9$
  - $p = \pm 3$

Examiner Tips and Tricks

Harder exam questions may not tell you to write $z$ as $x + y i$ (you have to spot it yourself).

Worked Example

Let $z$ and $w$ be complex numbers.
It is known that $w = 5 + 2 i$ and that $z^{2} + 10 w = z z *$ .

Find the two possible values of $z$ .

Write $z$ in the form $x + y i$

$z = x + y i$

Substitute this, and $w$ , into the equation

${(x + y i)}^{2} + 10 (5 + 2 i) = (x + y i) (x - y i)$

Expand the brackets and use $i^{2} = - 1$

$x^{2} + 2 x y i + y^{2} i^{2} + 50 + 20 i = x^{2} - y^{2} i^{2} x^{2} + 2 x y i - y^{2} + 50 + 20 i = x^{2} + y^{2}$

Equate the real parts and solve for $y$

$\begin{array}{rcl} x^{2} - y^{2} + 50 & = & x^{2} + y^{2} \\ 50 & = & 2 y^{2} \\ 25 & = & y^{2} \\ y & = & \pm 5 \end{array}$

Now equate the imaginary parts and solve for $x$
Note that there are no imaginary parts on the right

$\begin{array}{rcl} 2 x y + 20 & = & 0 \\ x y & = & - 10 \\ x & = & - \frac{10}{y} \end{array}$

When $y = 5$ , then $x = - 2$
When $y = - 5$ , then $x = 2$
Substitute these back in to get $z$

$- 2 + 5 i$ or $2 - 5 i$

Last updated: 10 April 2024

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Mark Curtis

Author: Mark Curtis

Expertise: Maths

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.