Let \(z = x + iy\) and \(z^* = x - iy\). Substitute into the equation:
\(\frac{5(x + iy)}{1 + 2i} - (x + iy)(x - iy) + 30 + 10i = 0\)
Multiply numerator and denominator by the conjugate of the denominator:
\(\frac{5(x + iy)(1 - 2i)}{(1 + 2i)(1 - 2i)} - (x^2 + y^2) + 30 + 10i = 0\)
\((1 + 2i)(1 - 2i) = 1 + 4 = 5\)
\(\frac{5(x + iy)(1 - 2i)}{5} - x^2 - y^2 + 30 + 10i = 0\)
\(5(x + iy)(1 - 2i) = 5x - 10y + i(5y + 10x)\)
\(x - 2ix + iy + 2y - x^2 - y^2 + 30 + 10i = 0\)
Equate real and imaginary parts:
Real: \(x + 2y - x^2 - y^2 + 30 = 0\)
Imaginary: \(-2x + y + 10 = 0\)
Solve the quadratic equations:
\(x^2 - 9x + 18 = 0\) gives \(x = 3\) or \(x = 6\)
\(y^2 + 2y - 8 = 0\) gives \(y = -4\) or \(y = 2\)
Thus, the solutions are \(3 - 4i\) and \(6 + 2i\).
