if j^2 = -1?

Question

what is the square root of j?

Answer 1

Recall that:

e^(j*theta) = cos(theta)+j*sin(theta)

so

e^(j*pi/2) = cos(pi/2)+j*sin(pi/2) = j

And so,

sqrt( j ) = j^(1/2) = [ e^(j*pi/2) ]^(1/2) = e^(j*pi/4)

However,

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

And that is the correct answer of sqrt(j). If you need some additional convincing, then multiply:

( sqrt(2)/2 + j*sqrt(2)/2 )^2 = 1/2 + j + j^2/2 = 1/2 + j - 1/2 = j

Clearly, (sqrt(2)/2 + j*sqrt(2)/2) = sqrt(j). In fact, you could have derived the answer this way by finding constants a and b that make (a + j*b)^2 = j true.

You could also check the Google calculator link below, where Google calculator says that sqrt(i) is 0.707106781 + 0.707106781 i, which agrees with everything above.

Answer 2

Square Root of J is 1

Answer 3

if j^2=-1 then j= to the square root of -1. but this is not true because radical -1 is imaginary so that means the square root of j is imaginary i.

Answer 4

Sorry .i only know about the square root of -1=i.

Answer 5

I haven't seen this before. Are you sure you want the square root of j or do you want to know what j itself is?

There is no "number" that when squared gives a negative answer, but there is the "imaginary number" denoted by the letter "i". i=sqrt(-1).
But if j^2= -1, and i=sqrt(-1), and j=sqrt(-1) then j=i and the square root of j = the sq rt of an imaginary number.
sqrt(j)=sqrt(i)=sqrt(sqrt(-1))

Answer 6

this is comutable and i shall calculate it

let (a+jb) be square root of j (in physics we use j as sqrt of -1 as i is for current)
a and b be rational

(a+jb)^2 = 0+1j
a^2-b^2+2jab = 0+ 1j

equating real part of both sides and imaginary of both sides
a^2= b^2 => a= +/-b (a=b or a =-b)
2ab = 1
put a =b we get 2a^2 = 1
a^2 = 1/2
a = 1/sqrt(2) = b=
a=b = sqrt(2)/2 or a=b = - sqrt(2)/2

a = -b is not possible because then 2ab is -2a^2 -ve. ( a and b are real

so square root of j = sqrt(2)(1+j)/2 or -ve of it

you can square both sides and check

Answer 7

square of any negative integer is imaginary and does not wxist so its imaginary.but for square root of 1 its +/-1(+ or - 1)

Answer 8

the -1 can't have a root number because it's a negative number.There isn't a number that multyplies to itself and give an negative number ex.1*1=1 so 1*?=-1 if ?=-1 then 1 isn't a root

Answer 9

This is an imaginary number. Anything squared is a positive number......

Answer 10

j^2 = -1
j^2 + 1 = 0
j^2 - i^2 = 0
(j+i)(j-i) = 0
j = i or j = -i
Take their square roots, you get 2 possible values of sqrt(j)
sqrt(i) or i^3/2