what is sq root of iota?

Answer 1

its 1

since i = (-1)^1/2

so i^2 = (-1)^(2*(1/2))

which is 1

Answer 2

Let x = sqrt(i) = sqrt[sqrt(-1)]
Now square both sides :
x^2 = sqrt(-1)
Now square both sides again :
x^4 = -1
or
x^4 + 1 = 0

This is a polynomial equation of the fourth degree
in which sqrt(i) is a zero.
In fact, there are 4 zeros to the equation.

To find them, note that they will all be imaginary.
So we can let x = a + bi

Thus, (a + bi)^4 = -1

Taking the square root of both sides gives :
(a + bi)^2 = ± sqrt(-1) = ± i

Expanding gives :
a^2 - b^2 + 2abi = 0 ± i

Equating real terms gives :
a^2 - b^2 = 0, so a^2 = b^2, or ± a = ± b,
so basically, we can just say that a = ± b.

Equating imaginary terms gives :
2ab = ± 1

Substituting either a = +b or a = -b gives : 2b^2 = ± 1
So, b^2 = ± 1/2, therefore, b = ± sqrt(2)/2 or ± i*sqrt(2)/2

Thus, there are 4 values for 'b' and 2 values for 'a'.
That makes 8 combinations possible for (a + bi), but
when the all the calculations are done, we find that
4 of the combinations are the same as the other 4.

The 4 values for sqrt(i) turn out to be :

sqrt(2)/2 + i*sqrt(2)/2
-sqrt(2)/2 + i*sqrt(2)/2
sqrt(2)/2 - i*sqrt(2)/2
-sqrt(2)/2 - i*sqrt(2)/2

or more succinctly : ± sqrt(2)/2 ± i*sqrt(2)/2.

Answer 3

There are two square roots of i (there are two square roots of any number)

they are:
√2/2 + (√2/2)i and
-√2/2 - (√2/2)i
or, expressed another way,
±√2(1+i)/2

(±√2(1+i)/2)^2 = (2/4)(1+i)^2 = (1/2)(1 + 2i + i^2) = (1/2)(2i) = i

Answer 4

I don't remember, but I will think it through.

i = i
i^2 = -1
I^3 = -i
i^4 = -1*-1 = 1
i^5 = i

What number if you squared it would give you i ?

I think you would have to leave it in the form of sqrt(i), it cannot be made any simpler, but I am inspired to read up on it.

Thanks for the question!

Answer 5

because | i | = 1

let square root of i = cos t + i sin t as r = 1

= e ^ it

so e^2it = i = e^ipi/2
so 2t = pi/2
t = pi/4 or pi/4+ pi/2

so we have square root = +/- (sqrt(2) + i sqrt(2))/2

Answer 6

(-1)^1/4