# Multiplication of Complex Numbers

Multiplying Complex Numbers

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• First, complex numbers can come in a variety of forms!

Ex: multiply $3 i \cdot - 4 i =$

Remember, with multiplication you can rearrange the order (called the Commutative Property):

$3 \cdot - 4 \cdot i \cdot i = - 12 {i}^{2}$

... and then always substitute -1 for ${i}^{2}$:

$- 12 \cdot - 1 = 12$

Ex: the numbers might come in a radical form:

$\sqrt{- 3} \cdot 4 \sqrt{- 12} =$

You should always "factor" out the imaginary part from the square roots like this:

$\sqrt{- 1} \sqrt{3} \cdot 4 \cdot \sqrt{- 1} \sqrt{4} \sqrt{3} =$

and simplify again:

$= i \cdot 4 \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{4}$
$= i \cdot 4 \cdot 3 \cdot 2 = 24 i$

Ex: what about the Distributive Property? $3 i \left(4 i - 6\right) =$

$= 12 {i}^{2} - 18 i$
$= 12 \left(- 1\right) - 18 i$
$= - 12 - 18 i$

And last but not least, a pair of binomials in a + bi form:

Ex: (3 - 2i)(4 + i) =

=12 + 3i - 8i - $2 {i}^{2}$
= 12 - 2(-1) + 3i - 8i
= 12 + 2 - 5i
= 14 - 5i

• Let ${z}_{1}$ and ${z}_{2}$ be two complex numbers.

By rewriting in exponential form,

$\left\{\begin{matrix}{z}_{1} = {r}_{1} {e}^{i {\theta}_{1}} \\ {z}_{2} = {r}_{2} {e}^{i {\theta}_{2}}\end{matrix}\right.$

So,

z_1 cdot z_2 =r_1e^{i theta_1}cdot r_2 e^{i theta_2} =(r_1 cdot r_2)e^{i(theta_1+theta_2)}

Hence, the product of two complex numbers can be geometrically interpreted as the combination of the product of their absolute values (${r}_{1} \cdot {r}_{2}$) and the sum of their angles (${\theta}_{1} + {\theta}_{2}$) as shown below.

I hope that this was clear.

• In trigonometric form, a complex number looks like this:
$a + b i = c \cdot c i s \left(\theta\right)$
where $a$, $b$ and $c$ are scalars.

Let two complex numbers:
$\to {k}_{1} = {c}_{1} \cdot c i s \left(\alpha\right)$
$\to {k}_{2} = {c}_{2} \cdot c i s \left(\beta\right)$
${k}_{1} \cdot {k}_{2} = {c}_{1} \cdot {c}_{2} \cdot c i s \left(\alpha\right) \cdot c i s \left(\beta\right) =$
$= {c}_{1} \cdot {c}_{2} \cdot \left(\cos \left(\alpha\right) + i \cdot \sin \left(\alpha\right)\right) \cdot \left(\cos \left(\beta\right) + i \cdot \sin \left(\beta\right)\right)$

This product will end up leading to the expression
${k}_{1} \cdot {k}_{2} =$
$= {c}_{1} \cdot {c}_{2} \cdot \left(\cos \left(\alpha + \beta\right) + i \cdot \sin \left(\alpha + \beta\right)\right) =$
$= {c}_{1} \cdot {c}_{2} \cdot c i s \left(\alpha + \beta\right)$

By analyzing the steps above, we can infer that, for having used generic terms ${c}_{1}$, ${c}_{2}$, $\alpha$ and $\beta$, the formula of the product of two complex numbers in trigonometric form is:
$\left({c}_{1} \cdot c i s \left(\alpha\right)\right) \cdot \left({c}_{2} \cdot c i s \left(\beta\right)\right) = {c}_{1} \cdot {c}_{2} \cdot c i s \left(\alpha + \beta\right)$

Hope it helps.

• To explain this, I will name two generic complex.
${c}_{1} = a \cdot c i s \left(\alpha\right)$ and ${c}_{2} = b \cdot c i s \left(\beta\right)$

The product between ${c}_{1}$ and ${c}_{2}$ is:
$a b \cdot c i s \left(\alpha\right) c i s \left(\beta\right) =$
$a b \cdot \left(\cos \left(\alpha\right) + i \sin \left(\alpha\right)\right) \left(\cos \left(\beta\right) + i \sin \left(\beta\right)\right) =$
ab*({cos(alpha)cos(beta)-sin(alpha)sin(beta)} +
$\left\{i \left(\sin \left(\alpha\right) \sin \left(\beta\right) + \cos \left(\alpha\right) \sin \left(\beta\right)\right\}\right) =$
$a b \cdot \left\{\cos \left(a + b\right) + i \sin \left(a + b\right)\right\}$//

Therefore, we can assume that the product of the two complex numbers ${c}_{1}$ and ${c}_{2}$ can be generaly given by the form above.

Ex.:
$\left(2 \cdot c i s \left(\pi\right)\right) \cdot \left(3 \cdot c i s \left(2 \pi\right)\right) = 6 \cdot c i s \left(3 \pi\right) = 6 \cdot c i s \left(\pi\right)$

Hope it helps.

• This key question hasn't been answered yet.

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