a² + b² + 2b + 4a + 5 = 0

We want to know what values a and b will have to be in order
for a and b to be real numbers and not imaginary numbers.

Write as a quadratic in "a":

a² + 4a + (b²+2b+5) = 0

In order to have real solutions, the discriminant of a 
quadratic must not be negative.

The discriminant is B²-4AC (Using capital letters to avoid
conflict of notation with small letters)

A = 1, B = 4, C = (b²+2b+5)

Discriminant = B²-4AC = 4²-4(1)(b²+2b+5) = 16-4(b²+2b+5) = 

16-4b²-8b-20 = -4b²-8b-4 = -4(b²+2b+1) = -4(b+1)²

This must not be negative, so the only value it can take on
so that -4(b+1)² is not negative is when it is zero.
which is when b+1=0 or b=-1, so b can only be -1.  So b=-1.

That makes the original equation:

a² + b² + 2b + 4a + 5 = 0

become

a² + (-1)² + 2(-1) + 4a + 5 = 0
        a² + 1 - 2 + 4a + 5 = 0
                a² + 4a + 4 = 0
                     (a+2)² = 0
                        a+2 = 0
                          a = -2

So since a=-2 and b=-1



Edwin

If  = , then (complete the squares (!) )

 = ,

which implies a = -2, b = -1  (if the sum of two squares of real numbers is zero, then both addends are zeros).

Having this, calculate the rest.