Question

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

Answer 1

See below.

Explanation:

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

`a+ib` where `i = sqrt(-1)` and where `a` and `b` are real numbers.

`bar(a+ib)=a-ib` is the known as the complex conjugate of `a+ib`.

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

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

`(a+ib)+(c+id) = (a+c) +i(b+d)`

Example, consider `(1+3i)+(2+i)`
Using the formula above we get: `(1+2) + i(3+1) = 3+4i`

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

`(a+ib)-(c+id) = (a-c) +i(b-d)`

Example, consider: `(1+3i)-(2+i)`
From this we get: `(1-2) + i(3-1) = -1+2i`

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

`(a+ib)(c+id)=ac+iad+ibc+i^2bd`
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:
`(a+ib)(c+id)=(ac-bd)+i(ad+bc)`

For example: `(1+3i)(2+i) = (2*1-3*1)+i(1*1+3*2)=-1+7i`

Division
This is a bit more complicated.

Remember above we stated that the complex conjugate of a complex number is: `bar(a+ib)=a-ib`, 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:

`(a+ib)(a-ib)=a^2+b^2+iab-iba`
`=a^2+b^2` This is a very important result that will allow us to do division.

So, given:

`(a+ib)/(c+id)`

To divide this we must multiply the fraction by the complex conjugate of the bottom of the fraction. That is `c-id`.

Now, `(c-id)/(c-id)=1` so in the following step we are simply multiplying be one and having no effect the on the value of the fraction:

`(a+ib)/(c+id)=(a+ib)/(c+id)*(c-id)/(c-id)`

Remember: `(a+ib)(a-ib)=a^2+b^2`, which we will apply to the bottom and the multiplication rule: `(a+ib)(c+id)=(ac-bd)+i(ad+bc)` which we will apply to the top to get:

`((ac-bd)+i(ad+bc))/(c^2+d^2)`

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

`(a+ib)/(c+id)=(ac-bd)/(c^2+d^2)+i(ad+bc)/(c^2+d^2)`

Let's try the example: `(1+3i)/(2+i)`

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

`(1+3i)/(2+i)= (1+3i)/(2+i)*(2-i)/(2-i)=(2+6i-i-3i^2)/(2^2+1^2) =(5+5i)/5`
`=1+i`

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