How do I find the sqrt(i)?

Question

How do I find the sqrt(i)?
I know the answer, but I need the workings.
Thanks

How do I find the sqrt(i)?
(i as in the imaginary number)
I know the answer, but I really need the workings.

Thanks

Answer 1

We'll assume it has to be some kind of complex number. So let a + bi = √i. Then squaring both sides gives you

(a + bi)^2 = i
a^2 + 2abi + (bi)^2 = i
a^2 + 2abi - b^2 = i
(a^2 - b^2) + 2abi = i
(a^2 - b^2) + 2abi = 0 + i

This means, if you break up the real and imaginary parts:
a^2 - b^2 = 0
2ab = 1

So a = b (we'll leave out the ± for now), and a = 1 / 2b
This means b = 1 / 2b, b^2 = 1/2, b = 1/√2, thus a = 1/√2

Now that we have a and b, the answer is (√2/2) + (√2/2)i. Though since the square root can be ± the root, you can take the negative of this too.

Answer 2

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[4] x^2+8x+25=0 Comparing the equation with standard quadratic equation ax^2+bx+c=0,we get a=1,b=8 and c=25 Therefore, x={-b+-sqrt(b^2-4ac)}/2a ={-8+-sqrt(64-100)}/2 =(-8+-sqrt(-36)/2 =(-8+-6 sqrti)/2 =-4+- 3 sqrt i,where i^2= -1

Answer 3

You want to consider Euler's representation for complex numbers. Every complex number can be written as having a magnitude and a phase:

z = R * e^(i theta)

where
R = sqrt(Re(z)^2 + Im(z)^2)
and
e^(i theta) = cos(theta) + i*sin(theta)

Doing things this way, you discover that
i = 1 * e^(i pi/2)
which means that the answer you're looking for is
sqrt(i) = sqrt(1) * sqrt(e^(i pi/2)) = 1 * e^(i pi/4) = cos(pi/4) + i sin (pi/4) = (1/sqrt(2))* (1+ i)

Answer 4

Sqrt I

Answer 5

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RE:
How do I find the sqrt(i)?
How do I find the sqrt(i)?
I know the answer, but I need the workings.
Thanks

Answer 6

This answer I found in

http://math2.org/math/oddsends/complexity/sqrti.htm

Rationale that ( i ) = (1/2) + i (1/2)

Assuming that
( i ) = (1/2) + i (1/2)
we can square both sides to get
i = [ (1/2) + i (1/2) ]2
i = [ (1/2) + 2(1/2) i + (1/2) i2 ]
i = [ (1/2) + i + (1/2) (-1) ]
i = i (which is a true statement)
This is not a proof, but simply evidence that the formula is correct.

Answer 7

i = cos (pi/2) + i sin (pi/2)

sqrt(i)
= [ cos (pi/2) + i sin (pi/2) ] ^ (1/2)
= cos (pi/4) + i sin (pi/4) [ De Moivre's Theorem ]

Statement of De Moivre's Theorem is
(cosx + i sinx) ^ n = cos(nx) + i sin(nx)

= [ 1/sqrt(2)] * ( 1 + i )

Answer 8

i is the square root of -1. And represents an imaginary number.

If we have +2, its square root is 1.414 approx.

If we have -2, its square root is 1.414 i

When we square 1.414 i, we get 2 x - 1 = -2

Answer 9

well i = sqrt(-1)..so sqrt(i)=fourthroot(-1)