INTEGRAL POWERS OF IOTA ( i )
In
fact we don’t have any solution to the given equation in the system of real
numbers.
Why? Because x2 + 1 = 0 gives x2 = -1
and square of every real number is non-negative.
Then,
what could we do?
We
need to extend the real number system to a larger system to find the solution
of the equation x2 + 1 = 0.
In general, we have to find the solution of the equation ax2
+ bx + c = 0,
where D = b2 – 4ac < 0.
Euler
was the first mathematician to introduce the symbol i ( Iota ) for the square root of -1, i.e. solution of the equation
x2 + 1 = 0.
Now,
we have
i
= √-1
i2
= -1
i3
= i2 * i = -i
i4
= ( i2 )2 = ( -1 )2 = 1
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