1. Finding the square root of a complex number
(a+bi)2 = ( a2 + b2) +2abi
so -2i = a + bi
Find -2i
Let -2i = ( a+bi)2
But a and b are REAL
-2i =( a2 + b2) +2abi
Equate Real Parts: 0 = a2 + b2
a = -1/b
Equate Im Parts: -2 = 2ab
0 = (-1/b)2 + b2
0 = 1/b2 + b2
0 = 1 + b4
(b2 - 1 ) ( b2 + 1 ) = 0
b2 =  1
b = +1, -1, or i

2. Solve the following equation (a+bi)2 = ( a2 + b2) +2abi
Solve z2 + 4z +(4+2i)=0
Z = -4   42 – [4x1x(4+2i)] 2
Z = -4   16 – i 2
Z = -4   – 8i =-2 -2i 2
-2i =( a2 + b2) +2abi
so b4 – 26b2 +16 = 0
Equate Real Parts: 0= a2 + b2
Equate Im Parts: -2 = 2ab
So b = +1, -1, or i
0 = (-1/b)2 + b2
a = -1/b
0 = 1/b2 + b2
So a = -1, +1, or i
0 = 1 + b4
-3 + i
Or i
(b2 - 1 ) ( b2 + 1 ) = 0
So z = i =
or –i =
b2 =  1