Find the square roots of the following:
(i) `7-24 i` (ii) `5+12 i`

Explanation

(i) Let `sqrt(7-42i) = x + iy ,` Then
`sqrt(7-24i) = x+iy`
` or 7-24i = (x+iy)^(2)`
`or 7 -24i =(x^(2) -y^(2)) + 2iy`
`or x^(2) - y^(2) = 7" "(1)`
` and 2xy = - 24" "(2)`
Now, `(x^(2)+y^(2) )^(2) = (x^(2)-y^(2))^(2)+4x^(2)y^(2)`
` or (x^(2)+y^(2))^(2) = 49+576= 625`
` or x^(2) +y^(2)" "[because x^(2)+y^(2) gt 0]" "(3)`
on solving (1) and (3), we get
`x^(2) = 16 and y^(2) = 9 rArr x = pm 4 and y = pm 3`
From (2), 2xy is negative. So, x and y are of opposite signs.
Hence, x =4 and y = - 3 or x= - 4 and y = 3
Hence, `sqrt(7-24i)= pm (4-3i)`
(ii) Let `sqrt(5+12i) = x + iy`, Then
`sqrt(5+12i) = x+iy`
`or 5+12i=(x+iy)^(2)`
`or 5+12i=(x^(2) -y^(2)) + 2ixy`
`or x^(2) - y^(2) = 5" (1)`
`and 2xy = 12" "(2)`
Now,`(x^(2) +y^(2))^(2)=(x^(2)-y^(2))^(2)+4x^(2)y^(2)`
`or (x^(2) +y^(2))^(2) = 5^(2) + 12^(2) = 169`
`or x^(2)+y^(2) = 13" "(because x^(2)+y^(2) gt 0)" (3)`
On sloving (1) and (2) , we get
`x^(2) = 9 and y^(2) = 4 rArr x = pm3 and y = pm 2`
From (2), 2xy is positive . So, x and y are of the same sign.
Hence,
`x = 3 and y = 2 or x = - 3 and y = -2`
Hence, `sqrt(5+12i) = pm (3+2i)`
(iii) Let `sqrt(-15 -8i) = x+iy`. Then
`sqrt(-15-8i) = x + iy`
` or -5-8i = (x+iy)^(2)`
`or -15-8i= (x+iy)^(2)`
`or -15 = x^(2) -y^(2)" "(1)`
`and 2xy = - 8 " "(2)`
Now `,(x^(2)+y^(2))^(2) = (x^(2) -y^(2))^(2) + 4x^(2) y^(2)`
`or (x^(2)+y^(2))^(2) = (-15)^(2) + 64 = 289`
`or x^(2)+y^(2) = 17`
On sloving (1) and (3), we get
`x^(2) = 1 and y^(2) = 16 rArr x = pm 1 and y = pm 4`
From (2), 2xy is negative . So, x and y are of oppsite signs.
Hence, x = 1 and y = - 4 or x = -1 and y = 4
Hence, `sqrt(-15-8i) =pm (1-4i)`.