What are complex numbers ?and explain me how to do the four arithmetical operations with them.

Feb 11, 2016

See below.

Explanation:

A complex number is any number which can be expressed as:

$a + i b$ where $i = \sqrt{- 1}$ and where $a$ and $b$ are real numbers.

$\overline{a + i b} = a - i b$ is the known as the complex conjugate of $a + i b$.

$a$ is the real part of the complex number and $b$ is the imaginary part.

In order to perform addition, add the real parts together, and add the imaginary parts together, like so:

$\left(a + i b\right) + \left(c + i d\right) = \left(a + c\right) + i \left(b + d\right)$

Example, consider $\left(1 + 3 i\right) + \left(2 + i\right)$
Using the formula above we get: $\left(1 + 2\right) + i \left(3 + 1\right) = 3 + 4 i$

Subtraction
To perform subtraction, simply subtract the real parts together and the imaginary parts together:

$\left(a + i b\right) - \left(c + i d\right) = \left(a - c\right) + i \left(b - d\right)$

Example, consider: $\left(1 + 3 i\right) - \left(2 + i\right)$
From this we get: $\left(1 - 2\right) + i \left(3 - 1\right) = - 1 + 2 i$

Multiplication
Complex numbers can be multiplied in a similar manner to expanding a pair of brackets:

$\left(a + i b\right) \left(c + i d\right) = a c + i a d + i b c + {i}^{2} b d$
Remember $i = \sqrt{- 1}$ so it follows that ${i}^{2} = - 1$. If we put this into the above and factor the $i$ out of the other terms we obtain:
$\left(a + i b\right) \left(c + i d\right) = \left(a c - b d\right) + i \left(a d + b c\right)$

For example: $\left(1 + 3 i\right) \left(2 + i\right) = \left(2 \cdot 1 - 3 \cdot 1\right) + i \left(1 \cdot 1 + 3 \cdot 2\right) = - 1 + 7 i$

Division
This is a bit more complicated.

Remember above we stated that the complex conjugate of a complex number is: $\overline{a + i b} = a - i b$, i.e. take the negative of the imaginary part.

Using multiplication from above: a complex number multiplied by it's own conjugate results in the following:

$\left(a + i b\right) \left(a - i b\right) = {a}^{2} + {b}^{2} + i a b - i b a$
$= {a}^{2} + {b}^{2}$ This is a very important result that will allow us to do division.

So, given:

$\frac{a + i b}{c + i d}$

To divide this we must multiply the fraction by the complex conjugate of the bottom of the fraction. That is $c - i d$.

Now, $\frac{c - i d}{c - i d} = 1$ so in the following step we are simply multiplying be one and having no effect the on the value of the fraction:

$\frac{a + i b}{c + i d} = \frac{a + i b}{c + i d} \cdot \frac{c - i d}{c - i d}$

Remember: $\left(a + i b\right) \left(a - i b\right) = {a}^{2} + {b}^{2}$, which we will apply to the bottom and the multiplication rule: $\left(a + i b\right) \left(c + i d\right) = \left(a c - b d\right) + i \left(a d + b c\right)$ which we will apply to the top to get:

$\frac{\left(a c - b d\right) + i \left(a d + b c\right)}{{c}^{2} + {d}^{2}}$

All is left is to split up the fraction and we get:

$\frac{a + i b}{c + i d} = \frac{a c - b d}{{c}^{2} + {d}^{2}} + i \frac{a d + b c}{{c}^{2} + {d}^{2}}$

Let's try the example: $\frac{1 + 3 i}{2 + i}$

First multiplying the top and the bottom by the conjugate of the bottom we get:

$\frac{1 + 3 i}{2 + i} = \frac{1 + 3 i}{2 + i} \cdot \frac{2 - i}{2 - i} = \frac{2 + 6 i - i - 3 {i}^{2}}{{2}^{2} + {1}^{2}} = \frac{5 + 5 i}{5}$
$= 1 + i$

Division is a bit intimidating at first glance but should come with practice.