find the square root of

- 5+12i
- -4-3i

**1).**

Let

⇒ 5 + 12 i = (x + iy)^{2}

⇒ 5 + 12 i = x^{2} - y^{2 }+ 2i xy

⇒ 5 + 12 i = (x^{2} - y^{2}) + 2i xy

⇒ x^{2} - y^{2 }= 5 ... (1)

and 2xy = 12 .... (2)

Now,

Solving equation (1) and (3),

*x*^{2} = 9 and *y*^{2} = 4

⇒ *x* = ± 3 and *y* = ± 2

From equation (2), 2*xy* is positve, so *x* and *y* are of the same sign.

∴ *x* = 3, *y* = 2 or *x* = – 3, *y* = -2

**2).**

Similarly, you can find out the square root of -4-3i.

**
**