square root of complex no.2

a)
Let z=x+iy

z=√(5+12i)
(x+iy)^2=5+12i
x^2+(2xy)i+(iy)^2=5+12i
(x^2-y^2)+(2xy)i=5+12i

By comparing, we have:
{x^2-y^2=5...(1)
{2xy=12 => y=6/x...(2)

Sub. (2) into (1):
x^2-(6/x)^2=5
x^2-36/x^2=5
x^4-5x^2-36=0
(x^2-9)(x^2+4)=0
x=3/-3 or x^2=-4 (rejected)

Hence, z=3+2i or z=-3-2i

b)
z^2+16-30i=0
z^2=-16+30i

Again, we let z=x+iy
(x+iy)^2=-16+30i
(x^2-y^2)+(2xy)i=-16+30i

By comparing, we have:
{x^2-y^2=-16...(1)
{2xy=30 => y=15/x...(2)

Sub. (2) into (1):
x^2-225/x^2=-16
x^4+16x^2-225=0
(x^2-9)(x^2+25)=0
x^2=9 or x^2=-25 (rejected)
x=3/-3

Hence, z=3+5i or z=-3-5i

c)
我諗你miss係話+/-4-/+3i等於+/-(4-3i)