Find all solutions to z^2 + 4conjugate[z] + 4 = 0 where z is a complex number.

let z = a + ib

you get (a+ib)^2 + 4(a-ib) + 4 = 0

expand

a^2 - b^2 + 2aib + 4a - 4bi + 4 = 0

(a^2 - b^2 + 4a + 4) + 2abi- 4bi = 0

equate imaginary and real parts on both sides to get

(a^2 - b^2 + 4a + 4) = 0 ...1
and 2abi - 4bi = 0 => b = 0 or a = 2

solve 1 using b = 0 to get a= - 2
sollve 1 using a= 2 to get b = +/-4

so (-2,0) is a solution so z = - 2
(2, 4i) and (2,-4i) are 2 other solutions so z = 2 + 4i or 2- 4i