 New Zealand
Level 8 - NCEA Level 3

Multiplication of Complex Numbers

Lesson

When multiplying complex numbers, we apply algebraic conventions as well as the fact that $i^2=-1$i2=1.

We can multiply two single terms together

• Real and Imaginary terms -> $3\times7i=21i$3×7i=21i
• Two imaginary terms -> $6i\times4i=24i^2=-24$6i×4i=24i2=24 (remember that $i^2=-1$i2=1)

We can a term with a complex number containing both real and imaginary parts

• Real number multiplied by a complex number $5\left(6+2i\right)=5\times\left(6+2i\right)=30+10i$5(6+2i)=5×(6+2i)=30+10i (just like expanding through brackets in algebra)
• Complex term multiplied by a complex number  $3i\left(-4+6i\right)=-12i+18i^2=-12i-18=18-12i$3i(4+6i)=12i+18i2=12i18=1812i

We can also undertake binomial expansion where the two bracketed values are complex numbers.

 $\left(1-3i\right)\left(4+2i\right)$(1−3i)(4+2i) $=$= $\left(1\right)\left(4\right)+\left(-3i\right)\left(4\right)+\left(1\right)\left(2i\right)+\left(-3i\right)\left(2i\right)$(1)(4)+(−3i)(4)+(1)(2i)+(−3i)(2i) $=$= $4-12i+2i-6i^2$4−12i+2i−6i2 $=$= $4-10i-6i^2$4−10i−6i2 $=$= $4-10i-6\left(-1\right)$4−10i−6(−1) $=$= $4-10i+6$4−10i+6 $=$= $10-10i$10−10i

and one more example

 $\left(4-3i\right)\left(4+3i\right)$(4−3i)(4+3i) $=$= $16-9i^2$16−9i2 $=$= $16+9$16+9 $=$= $25$25

This final case is a special example as the numbers are what we call conjugates of each other.  We will study more about these special cases later.

Activity

Try this yourself before checking out the solution

We already know the surd laws such as

$\sqrt{a}\times\sqrt{b}=\sqrt{ab}$a×b=ab

$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ab=ab

Extend these laws to see what happens if $a$a or $b$b or both $a$a&$b$b are <$0$0

(see here for the solution)

Worked Examples

QUESTION 1

Simplify $-6\left(3-5i\right)$6(35i).

QUESTION 2

Simplify $\sqrt{10}i\left(8+\sqrt{10}i\right)$10i(8+10i), writing your answer in terms of $i$i.

QUESTION 3

Simplify $\left(2+5i\right)\left(5i-2\right)$(2+5i)(5i2).

QUESTION 4

Simplify $-2i\left(4-3i\right)^2$2i(43i)2, leaving your answer in terms of $i$i.

Outcomes

M8-9

Manipulate complex numbers and present them graphically

91577

Apply the algebra of complex numbers in solving problems