Let f (z) = x ^ 2 + I * y ^ 2, then f '(1 + I) = result is 2. How to do it? Another problem is in the complex range, the number of roots of the equation Z ^ 3 + | Z | = 0? Let f (z) = x ^ 2 + I * y ^ 2, then f '(1 + I) = the result is 2. How do you do that? Another problem is that in the complex range, how many roots of the equation Z ^ 3 + | Z | = 0? The answer to the second question is four roots. I'm very tangled

Let f (z) = x ^ 2 + I * y ^ 2, then f '(1 + I) = result is 2. How to do it? Another problem is in the complex range, the number of roots of the equation Z ^ 3 + | Z | = 0? Let f (z) = x ^ 2 + I * y ^ 2, then f '(1 + I) = the result is 2. How do you do that? Another problem is that in the complex range, how many roots of the equation Z ^ 3 + | Z | = 0? The answer to the second question is four roots. I'm very tangled

(1) Taking z = x + iy as a binary function, f '(z) = DF / DZ = (DF / DX + DF / dy) / (DZ / DX + DZ / dy) = (2x + 2iy) / (1 + I) is substituted into f' (1 + I) = 2 (2) let z = x + iy, X and y be real numbers, then Z ^ 3 = (x + iy) ^ 3 = x ^ 3 + 3iyx ^ 2-3xy ^ 2-iy ^ 3, | Z | = x ^ 2 + y ^ 2, the original equation is transformed into (1) x ^ 3-3xy ^ 2 + x ^ 2 + y ^ 2 = 0 (2)