The Square of the Imaginary Unit $i$

In the following calculations, it is important to remember that $i^2=-1$. It is best to not let this fact confuse you, but to just remember it as a fact. Any time an $i^2$ appears in a calculation, it can be replaced by the real number $-1.$

Multiplying Complex Numbers

Two complex numbers can be multiplied to become another complex number. The key to performing the multiplication is to remember the acronym FOIL, which stands for First, Outer, Inner, Last. Thus, we multiply $a+bi$ and $c+di$ by writing $(a+bi)(c+di)$ and multiplying the First terms $a$ and $c$, and then the Outer terms $a$ and $di$ and then the Inner terms $bi$ and $c$ and then the Last terms $bi$ and $di$. Note that this last multiplication yields a real number, since:

$bi\cdot di = bd \cdot i^2=bd\cdot (-1) =-bd$ 

Note that the FOIL algorithm produces two real terms (from the First and Last multiplications) and two imaginary terms (from the Outer and Inner multiplications). We then combine these to write our complex number in standard form. Thus we have:

$(a+bi)(c+di)=ac+adi+bci-bd=(ac-bd)+(ad+bc)i$. 

For example, consider the product $(2+3i)(4+5i)$. 

We would compute:

 $\begin {aligned}(2+3i)(4+5i)&=8+10i+12i-15 \\&=(8-15)+(10+12)i \\&=-7+22i \end {aligned}$

As another example, consider the product:

 $\begin {aligned}(1-i)(2+4i)&=1\cdot 2 +1\cdot 4i -2 i -(4i^2) \\&= 2+2i-(-4) \\&=6+2i \end {aligned}$

Note that if a number has a real part of $0$, then the FOIL method is not necessary. For example: 

$\begin {aligned}(0+5i)(2+5i)&=5i(2+5i) \\&= 10i+25i^2 \\&=-25+10i \end {aligned}$

Similarly, a number with an imaginary part of $0$ is easily multiplied as this example shows: $(2+0i)(4-3i)=2(4-3i)=8-6i.$

Note that it is possible for two nonreal complex numbers to multiply together to be a real number. For example: 

$\begin {aligned} (2-3i)(2+3i)&=4+6i-6i-9i^2 \\&=4+9 \\&=13 \end {aligned}$