    # Square Root Of Complex Numbers TO FIND THE SQUARE ROOT OF THE COMPLEX NUMBER a + ib
WHERE a AND b (≠0) ARE REAL

Let √a + ib = x + iy where x and y are real.

Then, a + ib = (x + iy)2 = x2 + i2y2 + 2x. iy = x2 - y2 + i. 2xy

Equating real and imaginary parts we get,

x2 - y2 = a             (1)

2xy = b                 (2)

Now (x2 + y2)2 = (x2 - y2)2 + 4x2y2 = a2 + b2

Hence x2 + y2 = +√a2 + b2 (Since x, y both are real)             (3)

Now adding equations (1) and (3) we get, Now equation (3) - equation (1) gives, From equation (2) it is clear that both x and y will have the same signs (either both positive or both negative) when b is positive; and x and y will have different signs (one positive and the other negative) when b is negative.

Hence if b > 0 then the square roots of (a + ib) are: And if b < 0 then the square roots of (a + ib) are: Example 1:

Find the square root of Solution: Now let √1 + i = x + iy (where x and y are real)

Hence, 1 + i = (x + iy)2 = x2 - y2 + i. 2xy

Hence equating real and imaginary parts we get,

x2 - y2 = 1             (1)

2xy = 1                 (2)

Now (x2 + y2)2 = (x2 - y2)2 + 4x2y2 = 12 + 12 = 2

Or x2 + y2 = √2             (3)

Adding equations (1) and (3) we get, Now equation (3) - equation (1) gives, From (2) it is clear that xy > 0, hence x and y are both positive or both negative.

Hence the required square roots are: Online Solution Square Root of Complex Number Help:

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