Multiplying Complex Numbers – Techniques, Explanation, and Examples

Multiplying complex numbers is one of the most used operations involving complex numbers. The algebraic manipulations performed on complex numbers have similar processes when we algebraically manipulate binomials.

When multiplying complex numbers, we treat the imaginary and real number parts as two different variables.

We can then apply the different rules when we multiply two binomials.

In this article, we’ll explore all the possible techniques we might need when multiplying complex numbers. As mentioned, we might need to apply previous techniques we’ve learned when multiplying two binomials. If you need a quick refresher, check out these topics:

• Applying the perfect square trinomial
• Applying the difference of two squares
• Expanding binomials using the FOIL method.

Mastering these techniques will help us in becoming better at multiplying complex numbers. Why don’t we go ahead and see how we can apply these techniques when multiplying complex numbers?

How to multiply complex numbers?

Before we understand the different examples and techniques we can apply to multiply two complex numbers, let’s first refresh our knowledge on complex numbers.

Complex numbers are numbers that contain real and imaginary parts. In general, when we have $a + bi$, $a$ represents the real number part, and $bi$ represents the imaginary part.

Recall that $i = \sqrt{-1}$, so $i^2 = -1$. This is important to remember when multiplying two complex numbers. In general, here are the three guiding steps to remember:

1. Apply the different properties so we can remove any parentheses.
2. Replace $i^2$ with $-1$.
3. Combine real number and imaginary number parts to simplify.
4. Express the product in general form, $a + bi$.

Don’t worry. We’ll get into this in the next section and see how we can apply them in each case.

Complex number multiplication techniques

Let’s begin with the easiest case: when we distribute terms to simplify the product of complex numbers.

Applying the Distributive Property

Let’s say we want to multiply $i$ and $(4 – 2i)$. What we can do is multiply $i$ to each of the terms inside the parenthesis. Simplify the expression by using the fact that $i^2 = -1$.

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

Rearrange the terms to be written in general form, so the product is $4i + 2$ or $2 + 4i$.

Applying the Perfect Square Trinomial Property

When asked to find the product of a complex number and itself, we approach it like we would with binomials. Recall that $(m \pm n) = m^2 \pm 2mn + n^2$, and use this to expand the square of complex numbers of the form $a + bi$, where both $a$ and $b$ are not equal to $0$.

$ \begin{aligned}(3 – 4i)^2 &= (3)^2 – 2(3)(4i) + (4i)^2\\&=9 – 24i + 16i^2\end{aligned}$

Replace $i^2$ with $-1$ whenever possible and combine real and imaginary parts as well.

$ \begin{aligned}9 – 24i + 16i^2 &= 9 – 24i + 16(-1)\\&= 9 -24i – 16\\&= -7 -24i\end{aligned}$

Hence, $(3 – 4i)^2 = -7 – 24i$.

Multiplying Conjugates of Complex Numbers

In conjugate math, we learned to apply the difference of two squares when multiplying a complex number and its conjugate.

Let’s say, we have $a + bi$, so its conjugate is $a – bi$. Multiplying these two will return the following:

$\begin{aligned} (a – bi)(a + bi) &= (a)^2 -(bi)^2\\&= a^2 – b^2 i^2\\&=a^2 – b^2 (-1)\\&=a^2 + b^2\end{aligned}$

This means that to find the product of $2 + 3i$ and its conjugate, we simply square $2$ and $3$ then add the result.

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

Multiplying Complex Numbers Using the FOIL Method

We’ve established the fact that we treat complex numbers as binomials when finding their products, so the FOIL method will also apply.

As a refresher, FOIL stands for: “First,” “Inside,” “Outside,” and “Last.”  Let’s say we have two complex numbers, $(a + bi)$ and $(m + ni)$. We can apply the FOIL method to find their products, and below is the breakdown of how the FOIL method is applied.

$ \begin{aligned} (a + bi)(m + ni)\\\text{First}: a \cdot m \phantom{xx}\\\text{Outside}:a \cdot ni\\ \text{Inside}: bi \cdot m\phantom{x}\\\text{Last}: bi \cdot ni \phantom{xx}\end{aligned}$

Once we have these four pairs’ products, we can combine the real and imaginary parts to find the product of $(a + bi)$ and $(m + ni)$.

$\begin{aligned} (a + bi)(m + ni) &= am + ani + mbi + bni^2\\&=am + ani + mbi – bn\\&= (am -bn) + (an + mb)i\end{aligned}$

Hence, the general form of the product of two complex numbers, $(a + bi)$ and $(m + ni)$, is $(am -bn) + (an + mb)i$.

We can apply a similar process to find the product of $3 – 2i$ and $-2 + 4i$.

$\begin{aligned} (3 – 2i)(-2 + 4i)\\\text{First}: 3(-2) = -6 \\\text{Outside}: 3(4i) = 12i\\\text{Inside}: (-2i)(-2)= 4i\\\text{Last}: (-2i)(4i)= -8i^2\end{aligned}$

We’re listing down the pairs, but you don’t have to do it. You can even do it mentally if you can. This is just a guide in case you need a refresher.

$\begin{aligned} (3 – 2i)(-2 + 4i)&= -6 + 12i + 4i  -8i^2\\&=-6+12i + 4i +8\\&= (-6 + 8)+ (12 + 4)i\\&=2 + 16i\end{aligned}$

These are all the important concepts you may need to evaluate different expressions and equations involving the multiplication of complex numbers. The problems below will help us recall what we’ve just learned and apply these lessons to more complex problems.

Example 1

Evaluate and simplify the following expressions.

a. $-3(2 + 4i)$
b. $2i(-4 + 2i)$
c. $4i( 4 + 8i)$

Solution

Since the first factor only contains one term, we can distribute the term and multiply it with the terms inside the parenthesis.

Why don’t we start with $-3(2 + 4i)$?
$\begin{aligned} -3(2 + 4i)&= -3(2) -3(4i)\\&=-6 – 12i\end{aligned}$
a. This means that $-3(2 + 4i) = -6 -12i$.
Apply a similar process for $2i(-4 + 2i)$:
$\begin{aligned} 2i(-4 + 2i) &= 2i(-4) + 2i(2i)\\&=-8i +4i^2\end{aligned}$
We can further simplify the expressions by replacing $i$ with $-1$.
$\begin{aligned} -8i +4i^2 &= -8i + 4(-1)\\&= -8i – 4\\&= -4 -8i\end{aligned}$
b. Hence, $2i(-4 + 2i) = -4 -8i$.
Let’s distribute $4i$ and simplify the resulting expression by replacing $i^2$ with $-1$.
$\begin{aligned} 4i(4 + 8i)&= 4i(4) + 4i(8i)\\&=16i + 32i^2\\&= 16i + 32(-1)\\&=16i – 32\\&=-32 + 16i\end{aligned}$
c. The product of $4i( 4 + 8i)$ is $-32 + 16i$.

Example 2

Evaluate and simplify the following expressions using different algebraic properties.

a. $(-4 + 3i)(-4 – 3i)$
b. $(6 – 5i )(6 + 5i)$
c. $(-2 + 5i)^2$
d. $(\sqrt{3} – \sqrt{5}i)(\sqrt{3} – \sqrt{5}i)$

Solution

The first two expressions show two pairs of conjugates. Recall that given a pair of conjugates, $a – bi$ and $a + bi$, their product is equal to $a^2 + b^2$.

Let’s use this to find the product of $(-4 + 3i)$ and $(-4 – 3i)$.
$\begin{aligned} (-4 + 3i)(-4 – 3i)&= (-4)^2 + (3)^2\\&=16 + 9\\&= 25\end{aligned}$
a. From this, we can see that $(6 – 5i )(6 – 5i) = 25$.
We apply a similar process to find the product of $6 – 5i$ and $6 + 5i$.
$\begin{aligned} (6 – 5i)(6 + 5i) &= (6)^2 + (5)^2\\ &= 36 + 25\\&= 61\end{aligned}$
b. Hence, $(6 – 5i )(6 + 5i) = 61$.
When squaring a complex number, we can use the perfect square trinomial property, $(m \pm n)^2 = m^2 \pm 2mn + n^2$.
So, let’s go ahead and evaluate $(-2 + 5i)^2$ using this property.
$\begin{aligned} (-2 + 5i)^2 &= (-2)^2 + 2(-2)(5i)+ (5i)^2 \\&= 4 – 20i + 25i^2\end{aligned}$
Replace $i^2$ with $-1$ so that we can simplify the expression further.
$\begin{aligned}4 – 20i + 25i^2 &=4 – 20i + 25(-1)\\&=4 -20i – 25\\&=-21 – 20i \end{aligned}$
c. This means that $(-2 + 5i)^2$ is equal to $-21 -20i$.
Why don’t we go ahead and apply the same process for $(\sqrt{3} – \sqrt{5}i)(\sqrt{3} – \sqrt{5}i)$?
$\begin{aligned}(\sqrt{3} – \sqrt{5}i)(\sqrt{3} – \sqrt{5}i) &= (\sqrt{3})^2 – 2(\sqrt{5}i)(\sqrt{3})+ (\sqrt{5}i)^2\\&= 3 -2 \sqrt{5 \cdot 3}i + 5i^2\\&=3- 2\sqrt{15}i + 5(-1)\\&=3- 2\sqrt{15}i +-5\\&= (3-5) – 2\sqrt{15}i\\&= -2 – 2\sqrt{15}i\end{aligned}$
d. Hence, we have $-2 – 2\sqrt{15}i$.

Example 3

Evaluate and simplify the following expressions.

a. $(4 – 2i)(3 + i)$
b. $(-12 – 3i)(6 + 5i)$

Solution

To find the three pairs of complex numbers’ products, we’ll have to utilize the FOIL method.

Let’s begin by applying this on $(4 – 2i)(3 + i)$:
$\begin{aligned}(4- 2i)(3 + i)\\\text{First}: 4(3) = 12 \\\text{Outside}: 4(i) = 4i\\\text{Inside}: (-2i)(3)= -6i\\\text{Last}: (-2i)(i)= -2i^2\end{aligned}$
Once we product of each of the pairs of terms, let’s add these all up to find the product of $(4 – 2i)(3 + i)$. Make sure to replace $i^2$ with $-1$ to further simplify the result.
$\begin{aligned}(4- 2i)(3 + i) &= 12 + 4i – 6i -2i^2\\&=12 + 4i -6i + 2\\&=(12 + 2) + (4 -6)i\\&= 14 – 2i \end{aligned}$
a. This means that $(4 – 2i)(3 + i) = 14 -2i$.
Try to mentally write down the four resulting products of the terms when applying the FOIL method on $(-12 – 3i)(6 + 5i)$.
Don’t worry, we’ll show you what you should have so you can check your work as well:
$\begin{aligned}(-12 – 3i)(6 + 5i)\\\text{First}: -12(6) = -72 \\\text{Outside}: -12(5i) = -60i\\\text{Inside}: (-3i)(6)= -18i\\\text{Last}: (-3i)(5i)= -15i^2\end{aligned}$
Add these four “sub-products” to find the product of $(-12 – 3i)(6 + 5i)$.
$\begin{aligned}(-12 – 3i)(6 + 5i)&= -72 – 60i – 18i – 15i^2\\&=-72 – 60i – 18i + 15\\&=(-72 + 15) + (-60 -18)i\\&=-57 + (-78)i\\&=-57 – 78i\end{aligned}$
b. This means that $(-12 – 3i)(6 + 5i)$ is equal to $-57 – 78i$.

Example 4

Evaluate and simplify the following expressions.

a. $-2i(3 – 8i)(3 + 8i)$
b. $(5 – i)(2 + 4i)(4 – 3i)$
c. $(3 – 2i)^2(5 +12i)$

Solution

The three expressions will require us to use multiple techniques before we can evaluate and simplify the products.

It helps when you check for conjugates or perfect squares first since we have properties for their products.

The pair $3 – 8i$ and $3 + 8i$, for example, are conjugates, so it’ll be easier if we find their product first in $-2i(3 – 8i)(3 + 8i)$. Use the fact that $(a – bi)(a + bi) = a^2 + b^2$.
$\begin{aligned}-2i(3 – 8i)(3 + 8i)&= -2i[(3)^2 + (8i)^2]\\&=-2i(9 + 64i^2)\\&= -2i(9 – 64)\\&=-2i(-55)\end{aligned}$
We can then further simplify this by multiplying $-2i$ and $-55$.
$\begin{aligned}-2i(-55) &= (-2 \cdot -55)i\\&=110i\end{aligned}$
a. This means that $-2i(3 – 8i)(3 + 8i)$ is equal to $110i$.
There are no conjugates nor perfect squares in $(5 – i)(2 + 4i)(4 – 3i)$, so we can multiply the first two pairs of binomials first using the FOIL method.
$\begin{aligned}(5 – i)(2 + 4i)&= 5(2)+ 5(4i)+ 2(-i) + 4i(-i)\\&=10 + 20i -2i -4i^2\\&=10 + 20i -2i – 4(-1)\\&= 10 + 20i – 2i + 4\\&= (10 + 4) + (20 – 2)i\\&=14 + 18i\end{aligned}$
Let’s substitute this resulting product into our original expression then apply the FOIL method once more.
$\begin{aligned}(5 – i)(2 + 4i)(4 – 3i)&=(14 + 18i)(4 – 3i)\\&= 14(4) + 14(-3i) + 4(18i)+ 18i(-3i)\\&=56 – 42i + 72i- 54i^2\\&=56 – 42i + 72i – 54(-1)\\&= 56 – 42i+72i +54\\&=(56 +54) + (-42 + 72)i\\&= 110 + 30i\end{aligned}$
b. Hence, by using the FOIL method twice, we have $(5 – i)(2 + 4i)(4 – 3i) = 110 + 30i$.

For $(3 – 2i)^2(5 +12i)$, we can first expand $(3 – 2i)^2$ using the perfect square trinomial property, $(m \pm n)^2 = m^2 \pm 2mn + n^2$.
$\begin{aligned}(3 – 2i)^2 &= (3)^2 – 2(3)(2i) + (2i)^2\\&= 9 -12i + 4i^2\\&= 9 -12i + 4(-1)\\&= 9 -12i – 4\\&= (9 – 4) -12i\\&= 5 -12i\end{aligned}$
This means that we now have $(5-12i)(5 + 12i)$. These two are each other’s conjugate, so we can square both coefficients then add the result to find the final product.
$\begin{aligned}(5 -12i)(5 + 12i) &= (5)^2 + (12)^2\\&=25 + 144\\&= 169\end{aligned}$

c. This means that we have $(3 – 2i)^2(5 +12i) = 169$.