Final answer:
To prove that the function fₙ(z) = zₙ is analytic in ℕ, and to find its derivative, one may use the definition of complex derivatives or the Cauchy-Riemann equations. For all non-negative integers n, the function follows the rules of differentiation, making the derivative nfₙ₋₁(z).
Step-by-step explanation:
The student's question is about proving that the function fₙ(z) = zₙ is analytic in ℕ (the set of all complex numbers) and finding its complex derivative. This can be shown by either using the fundamental definition of the complex derivative or the Cauchy-Riemann equations. To prove that fₙ(z) is analytic, one could start by showing that its derivative exists at every point in ℕ. Since fₙ(z) is a polynomial function of z, it is differentiable, and thus, analytic in ℕ. It follows the algebraic rules of differentiation, so for n ∈ {0,1,2,…}, the derivative of fₙ(z) with respect to z would be nfₙ₋₁(z). This can also be verified using the Cauchy-Riemann equations, which state that if u(x,y) and v(x,y) are the real and imaginary parts of f(z) respectively, and both have continuous first partial derivatives which satisfy u_x = v_y and u_y = -v_x, then f(z) is analytic. In the case of fₙ(z), u and v will satisfy these conditions for all non-negative integers n.