Equating Real & Imaginary Parts (Edexcel IAL Further Maths)

Spec point — 1.1

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 $a+bi=8-10i$ 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 $z=x+yi$ for expressions in $z$

• For example, solve $z+2z\ast=12-i$

  • Let $z=x+yi$

  • Substitute in

    • $( x + y \mathrm{i} ) + 2 ( x - y \mathrm{i} ) = 12 - i$

  • Expand and collect terms

    • $3x-yi=12-i$

  • Equate real and imaginary parts

    • $3x=12$ 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 $z=x+yi$ then $\vert z \vert = \sqrt{x^{2} + y^{2}}$

• Helpful modulus rules are:

  • $\vert z_{1} z_{2} \vert = \vert z_{1} \vert \vert z_{2} \vert$

  • | z _ 1 / z _ 2 | = (| z _ 1 |)/(| z _ 2 |)

  • $\vert z \vert = \vert z \text{asterisk} \times \vert$

  • $z z \text{asterisk} \times = z \text{asterisk} \times z = \vert z \vert^{2}$

    • Proved using $z=x+yi$ and $z\ast=x-yi$

• For example, find $p$ if $z_{1}=p+\pi$, $z_{2}=2i$ and $\vert z_{1} z_{2} \vert = 6 \sqrt{2}$

  • Use $\vert z_{1} z_{2} \vert = \vert z_{1} \vert \vert z_{2} \vert$

    • $\vert p + p \mathrm{i} \vert \vert 2 \mathrm{i} \vert = 6 \sqrt{2}$

  • Use $\vert z \vert = \sqrt{x^{2} + y^{2}}$

    • $\sqrt{p^{2}+p^{2}}\times2=6\sqrt{2}$ so $2\sqrt{2p^{2}}=6\sqrt{2}$

  • Square both sides and simplify

    • $p^{2}=9$
    • $p=\pm3$

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

Worked Example Let $z$ and $w$ be complex numbers. It is known that $w=5+2i$ and that $z^{2}+10w=zz\ast$.

Find the two possible values of $z$.

Write $z$* ** in the form *$x+yi$

$z=x+yi$

Substitute this, and $w$, into the equation

• *$( x + y \mathrm{i} )^{2} + 10 ( 5 + 2 \mathrm{i} ) = ( x + y \mathrm{i} ) ( x - y \mathrm{i} )$

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

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

Equate the real parts and solve for $y$

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

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

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

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

Final answer: $-2+5i$or$2-5i$