Determine a complex number z=x+iy that satisfies the equation 3z -iZ = 16
Where z = complex number and Z=its conjugate.
Can't seem to crack the algebra in this one.. probably some obvious way to it that I missed no doubt..

Break the problem into real and imaginary parts, and then treat the problem as two equations in two unknowns. Remember that I*i = -1, and -i*i = +1.

3(x+iy) - I(x+iy) = 16+0i.
Real components: 3x +y = 16;
Imaginary components: 3y-x = 0.
Solve for x and y from there.

z = x + iy, Z = x - iy

3 (x + iy) - I (x - iy) = 16

3x + 3iy - ix - y = 16

3x - y + I(3y -x) = 16 + 0i

3x - y = 16, 3y -x = 0

Hence: y = 2 x = 6

z = 6 + 2i