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New Zealand
Level 8 - NCEA Level 3

Factorising and Expanding Complex Numbers


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+12i9
  $=$= $-5+12i$5+12i
Example 4:
$\left(1-i\right)^3$(1i)3 $=$= $1^3-3\left(1\right)^2\left(i\right)+3\left(1\right)\left(i\right)^2-i^3$133(1)2(i)+3(1)(i)2i3
  $=$= $1-3i-3+i$13i3+i
  $=$= $-2-2i$22i



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 $=$= $x^2-9i^2$x29i2
  $=$= $\left(x+3i\right)\left(x-3i\right)$(x+3i)(x3i)
Example 7:
$4z^2+25$4z2+25 $=$= $4z^2-25i^2$4z225i2
  $=$= $\left(2z+5i\right)\left(2z-5i\right)$(2z+5i)(2z5i)
Example 8:
$4x^2-12ix-9$4x212ix9 $=$= $4x^2-12i+9i^2$4x212i+9i2
  $=$= $\left(2x-3i\right)^2$(2x3i)2
Example 9:
$6z^2+iz+2$6z2+iz+2 $=$= $6z^2+iz-2i^2$6z2+iz2i2
  $=$= $\left(3z+2i\right)\left(2z-i\right)$(3z+2i)(2zi)




Manipulate complex numbers and present them graphically


Apply the algebra of complex numbers in solving problems

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