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

1 Answer
Feb 11, 2016

Answer:

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.