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
We can a term with a complex number containing both real and imaginary parts
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)
Simplify $-6\left(3-5i\right)$−6(3−5i).
Simplify $\sqrt{10}i\left(8+\sqrt{10}i\right)$√10i(8+√10i), writing your answer in terms of $i$i.
Simplify $\left(2+5i\right)\left(5i-2\right)$(2+5i)(5i−2).
Simplify $-2i\left(4-3i\right)^2$−2i(4−3i)2, leaving your answer in terms of $i$i.
Manipulate complex numbers and present them graphically
Apply the algebra of complex numbers in solving problems