The Square of any real number is positive.
Ex: \(2^{2}=4\)
\((-2)^{2}=4\)
\(( -\sqrt{3})^{2}=3\)
What are the solutions or roots of the following equations?
i) \(x^{2}-1=0\)
ii) \(x^{2}+1=0\)
Solution:
i) \(x^{2}-1=0\)
\(\Rightarrow\) (x-1)(x+1)=0
\(\Rightarrow\) (x-1)=0 or (x+1)=0
\(\therefore\) x=1, x=-1
The real solution of \(x^{2}-1\)=0 are 1, -1
ii) \(x^{2}+1=0\)
\(\Rightarrow\) \(x^{2}=-1\)
But we know squire of any real number is positive. That means it has no solution. So in real number system we can't solve \(x^{2}+1=0\)
So we extend Real number system to a large number system. This lage number system is called Complex number system.
In a complex number system, we are able to solve any equations like \(ax^{2}+bx+c=0\)
Where discriminant is positive negative or zero.
Mathematician Euler introduce the symbol iota (\(i\) for \(\sqrt{-1}\))
i.e, \((\sqrt{-1})^{2}\)=(\(i^{2}\))=-1
\(\sqrt{-4}\)=\(\sqrt{4}\)\(\sqrt{i}\)=(\2i\)
Now we can solve \(x^{2}+1=0\)
\(x^{2}\)+1=0\)
\(x^{2}=-1\)
\(x=\pm \sqrt{-1}\)
=\(\pm i\)
Defination of a complex number:
Anumber of the form x+iy where x and y are real numbers is called complex number. The set of cmplex number is denoted by 'C'.
a+ib \(\in\) C or C={(a, b): z=a+ib where z\(\in\)C and a\(\in\)R, b\(\in\)R}
Example:
2+3i is in the form a+ib
2+3i\(\equiv\)a+ib, a=2, b=3 \(\in\)R
z=a+ib=Re(z)+Im(z)
z has two parts. First one is real z and second one is imazinary z.