What is the square root of i?

Question

i = sqrt(-1)

Answer 1

Let the square root of i be (a+ib) where a,b are real numbers
(a+ib)^2 = i
a^2 - b^2 + 2abi = i

a^2-b^2=0
2ab=1

b=1/(2a)=(1/2)/a

a^2 - 1/4a^2 = 0
a^4 - 1/4 = 0
a=±1/sqrt(2)
b = (1/2)/a = ±1/sqrt(2)

sqrt(i) = ±(1/sqrt(2))(1+i)

Answer 2

i => imaginary part of a complex number.

Complex numbers can be expressed in Euler's form as

r*e^(i*theta) = r(cos(theta) + i*sin(theta)) - [1]

If r = 1 and theta = pi/2 = 90 degrees, we get

cos(pi/2) + i*sin(pi/2)
= 0 + i (since cos(pi/2) = 0 and sin(pi/2) = 1)
= i

In Euler's form, i = e^(i*pi/2). Taking square root on both sides we get,

sqrt(i) = i^(1/2) = (e^(i*pi/2))^(1/2) = e^(i*pi/4).

Expanding the Euler form as specified in [1] we get,

sqrt(i) = i^(1/2) = e^(i*pi/4) = cos(pi/4) + i*sin(pi/4)

we know that cos(pi/4) = sin(pi/4) = 1/sqrt(2) = 0.707106781

Therefore,

sqrt(i) = i^(1/2) = 0.707106781 + 0.707106781 * i

which is again a complex number.

You can also compute the square root of i by typing "square root of i" in the search text box of www.google.com and it will give you the same answer. Explanation above is the math on how you go about calculating it by yourself.

Answer 3

The sqrt of -1 is -1.
Because -1X-1=-1

Answer 4

i = exp(i*pi/2)

[exp(i*pi/2) = cos(pi/2) + i*sin(pi/2)]

sqrt(i) = exp(i*pi/4) = sqrt(2)/2 + i*sqrt(2)/2

Technically, we would also admit the negative root too:

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

But there's more. Since exp is periodic, the roots would be:

sqrt(i) = exp(i*pi/4 + 2*pi*n/2) = exp(i*(4n + 1)*pi/4), for all n = integers.

They are numerically equal. So the two roots are as above.

Answer 5

In polar form, i = cos (pi/2) + i sin(pi/2). So, according to Moivre's formula, i has 2 square roots,

i1 = cos(pi/4) + isin(pi/4) and
i2 = cos(5pi/4) + i sin(5pi/4). Thse roots are symmetric.

In algebraic form,

i1 = sqrt(2)/2 + sqrt(2)/2 i

i2 = -sqrt(2)/2 - sqrt(2)/2 i

Answer 6

i is imaginary (does not exist) so it's square root also does not exist

but lets pretend

i=sqrt (-1) sqrt(i) = sqrt ( sqrt (-1))

but your stuck there..because sqrt -1 is imaginary
so if you continue by converting the squares into exponents you end up getting a false answer.

Answer 7

This is why I'm not a mathematician.. I get a headache trying to solve stuff like this.

Answer 8

Not possible. i itself is imaginary.