Equating Real & Imaginary Parts (Edexcel International A Level Further Maths)

Revision Note

Author

Mark Curtis

Expertise

Maths

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
  - $p^{2} = 9$
  - $p = \pm 3$

Exam Tip

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 $2 - 5 i$

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