89.9k views
5 votes
Consider the function fₙ (z)=zⁿ

(a) For n∈{0,1,2,…}, use either the fundamental definition of the complex derivative or the Cauchy-Riemann equations to prove that fₙ(z) is analytic in C. Also, derive its complex derivative.

User Robooneus
by
8.9k points

1 Answer

6 votes

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.

User Nicolas Gehlert
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories