how would u derive the square root of i?

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)

x = i^(1/2)

-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.

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.

Take it in polar form

i= 1<pi/2 +2kpi so sqrt(i) = 1<pi/4 +kpi ( k=0 and 1) =1<pi/4 and

1<5pi/4

so in binomial form

sqrt(i) =+-sqrt(2) /2 ( 1+i)

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)

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

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

.