how would u derive the square root of i?

Question

equation: x^2 = i

Answer 1

x = a+bi
x^2 = a^2 + 2abi - b^2 = i

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

a = b or a =-b
in First case a = b = 1/sqrt(2)
in second case a = 1/sqrt(2) and b = -1/sqrt(2)

Square roots of i are
x = (1+i)/sqrt(2) and
x = (1-i)/sqrt(2)

Answer 2

x = i^(1/2)

Answer 3

-1 = i * i
so
i = sqrt(-1)

therefore
-1 = j * j * j *j
or
j = sqrt(sqrt(-1))

the value "i" is very very useful, and pops up everywhere in sciences. Its on the same scale of value as pi, and a little less useful than e.

Fields that its useful in:
heat transfer
elasticisty
fluid flow
acoustics
structural dynamics
optics
ac electronic circuits
SISO/LTI control systems
MIMO/not-LTI control systems

Although you could find nth roots of negative 1 you might not find another so useful.

To find another so useful its probably wise to first understand how i is used in each of the above fields and what properties make it so useful.

Answer 4

You do not, as it might seem, need a new set of imaginary units to obtain the squareroot of an imaginary number.

For i, it is +/- 1/sqrt(2) (1 + i)^2

Expanding the square of the expression proves that it is the squareroot of i.

Answer 5

Take it in polar form
i= 1 1<5pi/4
so in binomial form
sqrt(i) =+-sqrt(2) /2 ( 1+i)

Answer 6

sqrt(i) = (a+bi) ----- a and b are real numbers
(a+bi)² = i
a² + 2abi - b² = i
a² - b² = 0
a = b
or a = -b

a=b
2abi = i
2a² = 1
a = 1/sqrt(2)
b = 1/sqrt(2)

sqrt(i) = ±(1+i)/sqrt(2)

a = -b
2abi = i
-2a² = 1
a = sqrt(-1/2) ---- not real


i has 2 square roots:
±(1+i)/sqrt(2)

Answer 7

i is an imaginary number that is defined as the square root of -1.

Answer 8

de moivres thereom backwards
i = e^iπ/2
so
√i = e^i(π/2 + 2πk)/2 = e^i(π/4 + πk) for k = 0,1

= e^iπ/4 = cos(π/4) + isin(π/4) = 1/√2 + i1/√2
and
= e^i(π/4 + π)= e^i5π/4 = cos(5π/4) + isin(5π/4) =1/√2-i1/√2
so
√i = 1/√2 ± i1/√2

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