NZ Level 8 (NZC) Level 3 (NCEA) [In development] Factorising and Expanding Complex Numbers
Lesson

## Simplifying powers of complex numbers

The expression $(a+bi)^n$(a+bi)n, for non-negative integers $n$n, is expanded as any other binomial expansion, and then simplified by recalling that integer powers of $i$i become either $1$1, $-1$1, $i$i or $-i$i.

Here are some examples:

##### Example 1:

Expand and simplify $9i\left(x+7i\right)$9i(x+7i), leaving your answer in terms of $i$i.

##### Example 2:

Expand and simplify $\left(7+2i\right)\left(7-2i\right)$(7+2i)(72i).

##### Example 3:
 $\left(2+3i\right)^2$(2+3i)2 $=$= $4+12i+9i^2$4+12i+9i2 $=$= $4+12i-9$4+12i−9 $=$= $-5+12i$−5+12i
##### Example 4:
 $\left(1-i\right)^3$(1−i)3 $=$= $1^3-3\left(1\right)^2\left(i\right)+3\left(1\right)\left(i\right)^2-i^3$13−3(1)2(i)+3(1)(i)2−i3 $=$= $1-3i-3+i$1−3i−3+i $=$= $-2-2i$−2−2i

## Factoring

With factorisation, the key is to introduce the term $i^2=-1$i2=1 in the given expression whenever required. Here are a few examples:

##### Example 5:

Factorise the expression $3x^2+108$3x2+108. Leave your answer in terms of $i$i.

##### Example 6:
 $x^2+9$x2+9 $=$= $x^2-9i^2$x2−9i2 $=$= $\left(x+3i\right)\left(x-3i\right)$(x+3i)(x−3i)
##### Example 7:
 $4z^2+25$4z2+25 $=$= $4z^2-25i^2$4z2−25i2 $=$= $\left(2z+5i\right)\left(2z-5i\right)$(2z+5i)(2z−5i)
##### Example 8:
 $4x^2-12ix-9$4x2−12ix−9 $=$= $4x^2-12i+9i^2$4x2−12i+9i2 $=$= $\left(2x-3i\right)^2$(2x−3i)2
##### Example 9:
 $6z^2+iz+2$6z2+iz+2 $=$= $6z^2+iz-2i^2$6z2+iz−2i2 $=$= $\left(3z+2i\right)\left(2z-i\right)$(3z+2i)(2z−i)

### Outcomes

#### M8-9

Manipulate complex numbers and present them graphically

#### 91577

Apply the algebra of complex numbers in solving problems