Given,

X = {z C : Re(az 2 + bz)= a and Re(bz 2 + az)= b}

Now, let z = x + iy

So, Re (az 2 + bz) = a

Ref a (x + iy) 2 + b(x+iy) = a

Re (a (x 2 - y 2 + 2ixy) + b (x + iy)) = a

a (x 2 - y 2 ) + bx = a ... (i)

And, Re(bz 2 +az) = b

Re (b (x + iy) 2 + a (x + iy)) = b

Re (b (x 2 - y 2 + 2ixy) + a (x + iy) = b

b (x 2 -y 2 ) + ax = b ... (ii)

From (i) and (ii), (i) -(ii)

we get, (x 2 - y 2 ) (a - b)-x (a - b) = a - b

X 2 – y 2 - X = 1 . . . (iii)

From (i) and (ii), (i)+(ii)

((x 2 - y 2 )+x - l) (a + b)= 0 (Note: here a + b 0 is considered but it is not clear from the question)

x 2 - y 2 + x - 1 = 0 ... (iv)

Now from (iii) and (iv) we get,

x = 0, y 2 = -1 (No solution)

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