Multiplicação entre números complexos

$$\text{**Multiplying Complex Numbers**}$$

Before we start, as a reminder, $i$ is an imaginary number!

When you multiply two complex numbers in rectangular form, you can use the FOIL method to get your final answer. In other words, for two complex numbers $z_1=a_1+ib_1$ and $z_2=a_2+ib_2,$ $$\begin{align} z_1z_2 &= (a_1+ib_1)(a_2+ib_2)\\ &= a_1a_2+a_1ib_2+ib_1a_2+i^2b_1b_2\\ &=(a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2) \end{align}$$

Remember, it's important to simplify the powers of $i.$ $i$ makes up the imaginary parts!

Let's do a simple example to practice!

To solve this problem, we use the FOIL method on $z_1=1+i$ and $z_2=3+2i.$ $$\begin{align} z_1z_2 &= (1+i)(3+2i)\\ &= 3+2i+3i+2i^2\\ &= 3+2i+3i-2\\ &= 1+5i\\ \end{align}$$ Notice how we substituted in $-1$ wherever there was an $i^2.$ This trick is extremely helpful in simplifying our final answer.

We can also multiply complex numbers when they are in polar form! Sometimes, this is a more efficient method, with less FOILing and algebraic manipulation.

In a moment we're going to be talking about $i$, but before that let's start off with a very easy question.

What happens when you multiply a number by $-1$?

When you take a number like $5$ and multiply it by $-1$, you get $-5$. Let's think about what change happens when you do that.

A picture might clear things up. Take a look.

So multiplying something by $-1$ rotates that number by $180^\circ$ on the number line.

Now you can imagine what happens when you multiply that number by $-1$ again.

It gets rotated by $180^\circ$ again doing a full rotation and going back to where it started from!

So this is a great way to visualize why $-(-x)=x$.

But let's not get off-topic. We're supposed to talk about $i$ here.

Well, let's get to it!

Now think of a number $k$ such that when you multiply any number by it, it gets rotated by $90^\circ$. You should have no trouble with this I hope. After all we just saw a number that rotates everything by $180^\circ$. So a $90^\circ$ -rotator number shouldn't be that hard.

Now what happens when you multiply a number (say $5$) by $k$ twice?

Well, doing it once rotates the number by $90^\circ$. So doing it twice should rotate the number by $90^\circ+90^\circ=180^\circ$. And a rotation by $180^\circ$ is equivalent to a multiplication by $-1$.

In other words,

$$5\times k^2=5\times(-1)$$

So, $$k^2=-1$$.

This number $k$ is what everyone calls $i$. This is the so-called imaginary unit. Surely it's not a real number. But it's no more imaginary than $-1$.

Simplifying expressions with $i$ involves using patterns in powers of imaginary numbers.

That's all for now! I hope you had fun learning this.

Until next time!

On the complex, multiplication of complex numbers extremely similar to binomial expansion (FOIL). Just treat $i$ as a variable at first. $$(3+4i)(2+5i)$$ $$3(2+5i) + 4i(2+5i)$$ $$6+15i+8i+20i^2$$ However, the last crucial step occurs with the property that $i^2=-1$: we replace every $i^2$ with -1. $$6+23i-20$$ $$-14+23i$$

Multiplication of more complex numbers can be done by multiplying the product of the first two with the third, and so on.

(Video) Multiplying Complex Numbers

by mathman1024

Here's a video by mathman1024 where he shows you how to multiply two complex numbers together.

The first example he does is: $(2+9i)(3-5i)$. To multiply these we need to use the FOIL method. $$\begin{align} &(2+9i)(3-5i)\\ =&6-10i+27i-45i^2\\ =&6-10i+27i-45\color{#8925AE}{(-1)}\\ =&6-10i+27i\color{#8925AE}{+}45\\ =&51+17i \end{align}$$

Try the next one on your own and see if you get the same answer.

1. $(5-3i)(6-4i)$

Multiplying Complex Numbers

A complex number is written in the form: $a + bi$

, where $a$ is the real part and $b$ is the imaginary part.

To multiply complex numbers, you use the same method as multiplying binomials, which is the distributive property. That's FOIL! The most important rule to remember is:

$$i^2=-1$$

Let's look at an example:

$$(2+3i)(4+5i)$$

First, multiply each part:

$$2 \cdot 4 = 8$$ $$2 \cdot 5i = 10i$$ $$3i \cdot 4 = 12i$$ $$3i \cdot 5i = 15i^2$$

Now combine everything:

$$8+ 10i + 12i + 15i^2$$

Combine like terms:

$$8 + 22i + 15i^2$$

We know that $i^2 = -1$, so we can rewrite:

$$15i^2 = 15(-1) = -15$$

Now simplify:

$$8-15+22i = -7 + 22i$$

So the final answer is:

$$-7+22i$$

Multiplying complex numbers is just like regular multiplication, but don't forget to simplify using $i^2=-1$!

In future lessons, you will also learn how to dividing complex numbers. You can also extend this topic to multiplying complex numbers in polar form!