解方程..............

b + [6b / (a2 + b2)] = 0

b[1 + 6 / (a2 + b2)] = 0

b = 0 or 1 + 6 / (a2 + b2) = 0

b = 0 or a2 + b2 = - 6 (rejected) (Since a and b are real numbers)

For b = 0,

a – [6a / (a2 + 02)] = 1

a – 6a / a2 = 1

a – 6 / a = 1

a2 – a – 6 = 0

(a - 3)(a + 2) = 0

a = 3 or -2

So, a =-2, b = 0 or a = 3, b = 0

a-[(6a)/(a^2+b^2)]=1 ... (1)
b+[(6b)/(a^2+b^2)]=0 ...(2)
From (2), b = -6b/(a^2+b^2)
b(a^2+b^2) = -6b
b[a^2+b^2 + 6] = 0
b = 0 or a^2 + b^2 + 6 = 0 (rejected, since a^2, b^2>=0 for real values of a and b, hence "a^2+b^2 not equal to -6" for any a and b.)
Substitute b = 0 into (1)
a- 6a/(a^2+0) = 1
a-6/a = 1
a^2 - a - 6 = 0
(a-3)(a+2) = 0
a = 3 or -2
Hence, a = 3 and b = 0 OR a = -2 and b = 0