# Square Root Of Complex Numbers

TO FIND THE SQUARE ROOT OF THE COMPLEX NUMBER a + ib

WHERE a AND b (≠0) ARE REAL

WHERE a AND b (≠0) ARE REAL

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

Then, a + ib = (x + iy)

^{2}= x^{2}+ i^{2}y^{2}+ 2x. iy = x^{2}- y^{2}+ i. 2xy
Equating real and imaginary parts we get,

x

^{2}- y^{2}= a (1)
2xy = b (2)

Now (x

^{2}+ y^{2})2 = (x^{2}- y^{2})2 + 4x^{2}y^{2}= a^{2}+ b^{2}
Hence x

^{2}+ y^{2}= +√a^{2}+ b^{2}(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}= x^{2}- y^{2}+ i. 2xy
Hence equating real and imaginary parts we get,

x

^{2}- y^{2}= 1 (1)
2xy = 1 (2)

Now (x

^{2}+ y^{2})^{2}= (x^{2}- y^{2})^{2}+ 4x^{2}y^{2}= 1^{2}+ 1^{2}= 2
Or x

^{2}+ y^{2}= √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:

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