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:

 

Example 2:

 

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:

 

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)