Multiplication of Complex Numbers
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=−12i−18=−18−12i
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.
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
$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$√a√b=√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(3−5i).
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)(5i−2).
QUESTION 4
Simplify $-2i\left(4-3i\right)^2$−2i(4−3i)2, leaving your answer in terms of $i$i.