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

Hey guys,

my only problem with this question is how i deal with the complex conjugate here.

any help would be greatly appreciated!!

Hello,

You define a and b, the respective real and imaginary parts of z and express the equation with them.

= = = = = = = = = = = = = = = = = = = = = = = = = = =
z² + 4z* + 4 = 0
(a + ib)² + 4(a - ib) + 4 = 0
a² + 2iab - b² + 4a - 4ib + 4 = 0
(a² - b² + 4a + 4) + i(2ab - 4b) = 0

So by equating real and imaginary parts to 0, you obtain the system:
{ b² = a² + 4a + 4 = (a + 2)²
{ b(a - 2) = 0

Using the second equality, either b=0 and thus, through first equality, a=-2,
or a=2 and thus, through first equality, b²=16 yielding b=±4

Hence the solutions set is
{ -2; 2-4i; 2+4i }

Regards,
Dragon.Jade :-)