proof for imaginary number problem?

Question

j is an imaginary number i.e the sq root of -1 (it used to be denoted as "i" but "j" is more used now a days.
i was told sqrroot j or i = (1+j)/2



can some one prove this..?

Answer 1

The easiest way by far to do roots and such of complex numbers is to use polar notation. You can visualize polar notation as taking a number z = x + i*y and plotting the real part (x) along the x axis and the imaginary part (y) along the y-axis. The complex number can also be expressed a z = R * e^(i*a). Using Euler's identity:

e^(i * a) = cos(a) + i * sin(a)

It is pretty easy to show that: R = sqrt(x^2 + y^2) which is the length of the line on the polar plot and a = atan(y/x) which is the angle the line makes with the x-axis. Of course, you do need to be careful to get the arctangent in the right quadrant,

Once you have converted to polar form, the square root becomes easy since taking the root of an exponent just divides the exponent by two:

sqrt(R * e^(i * a)) = sqrt(R) * sqrt(e^(i * a)) = sqrt(R) * e^(i*a/2)

Something interesting happens in polar form too. I can add 2*pi to the angle and my complex number remains unchanged:

R* e^(i * a) = R*e^(i*(a+2*pi))

If I take the square root of the second form, i get sqrt(R) * e^(i*(pi + a/2))

There are 2 distinct square roots of any number. If I add 2*pi again and take the root, I will end up with the first root again. If you carry this farther, you find there are n distinct roots for z^(1/n) when n is an integer.

Enough instruction. On to your problem. I will find both roots of i. Since i is just a vertical line of length 1on the polar plot it has a magnitude of 1 and an argument (or angle) of pi/2:

i = 1*e^(pi/2)

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

The second root:

sqrt(i) = sqrt(1) * e^(pi/4 + 2*pi) = cos(5*pi/4) + i*sin(5*pi/4) = -(1 + i)/sqrt(2)

For curiosity, just what is the square of (1 + i)/2?

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

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

So the the proper identity could be either:

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

or:

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

The important thing is to remember to us polar notation on this type of probelm. The problem solution becomes almost trivial once you understand the concept.

Finally, when I saw you problem statement, I solved it by visualizing the plot (a vertical line of unit length) and did the square root by taking the square root of the length (1) and splitting the angle of pi/2 in half. I immediately saw the square root of i to be e^(pi/4).

Answer 2

"j" is not used more often. It is just used by certain groups that use "i" for other things (like electrical engineers).

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). Note that your answer is not correct.

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). 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 3

let √i =a+bi where a and b are real.
then i=(a+bi)^2
which is a^2-b^2+2abi
therefore a^2=b^2 and 2ab=1
or a=plus or - b
ab=1/2
a^2=1/2
a=√1/2 and b=√1/2
or a=-√1/2 and b=-√1/2
a and b would not be real
√I = plus or minus √1/2(1+i)
Nb using the imaginary solution would give you the same solution
your solution is not correct