how can I find the square root of i (imaginary number)?

Answer 1

Let the square root of i be (a+ib) ---- a and b are real numbers.

(a+ib)² = i

Expand:
a² + 2abi - b² = i
a² - b² = 0

a = b
OR
a = -b

2ab = 1
If a=b
2a² = 1
a = 1/√2
b = 1/√2

If a=-b
-2a² = 1
a is not real, which violates our initial assumption.

sqrt(i) = ±(1/√2 + i/√2)
= (±1/√2)*(1 + i)

Answer 2

Well, you can "imagine" (no pun intended) that sqrt(i) is some complex number: call it a+bi. Our job is to find what the the values of "a" and "b" are.

Since a+bi = sqrt(i), we have:

(a+bi)(a+bi) = i
a² + 2abi + (bi)² = i
a² + 2abi + b²i² = i
a² + 2abi - b² = i
(a² - b²) + 2abi = i

Now, the "real" part of i is zero, so:
(a² - b²) = 0
or:
a² = b²

And the "imaginary" part if i is 1: so
2ab = 1
or:
ab = 1/2

If you solve those two equations simultaneously, you get two solutions:

solution 1:
a = 1/sqrt(2); b = 1/sqrt(2)

solution 2:
a = −1/sqrt(2); b = −1/sqrt(2)

So, this means the two square roots of i are:

1/sqrt(2) +(1/sqrt(2))(i)
and:
−1/sqrt(2) − (1/sqrt(2))(i)

Answer 3

In polar form i=1 For k=0 1 and for k=1 =-sqrt(2)/2*(1+i)

Answer 4

There are 2 square roots of i:
(1/sqrt2)(1+i) &
(-1/sqrt2)(1+i)

Answer 5

i is the sqrt of -1 [or (-1)^-2]
So, the sqrt of i must be (-1)^-4

Answer 6

√i

√(4i) = 2√i

√(8i) = 2√(2i)

:)