Question

Consider the function f(z) = z, where z E C. a). Use the epsilon to delta definition of limit to show that f is continuous for all z = z₀. We will now show that g(z) = zⁿ is continuous at z = z₀ using the method of induction. First, we establish a base case for n=1. That is, we demonstrate that the function f(z) = z is continuous at z = z₀ (notice that this was accomplished in part (a)).

Answer 1

Yes, the function
`\( f(z) = z \) is continuous for all \( z = z_0 \)`.

Step-by-step explanation

In the epsilon-delta definition of continuity, we aim to show that for any given epsilon (ε), there exists a delta (δ) such that whenever
`\( |z - z_0| < \delta \), it implies \( |f(z) - f(z_0)| < \epsilon \)`. For the function \
`( f(z) = z \)`, when
`\( z = z_0 \), \( |f(z) - f(z_0)| = |z - z_0| \).` Therefore, choosing
`\( \delta = \epsilon \) suffices, as it ensures that \( |z - z_0| < \delta \) implies \( |z - z_0| < \epsilon \). Hence, the function \( f(z) = z \) is continuous at any point \( z = z_0 \).`

Now, for the induction step, let's consider the function
`\( g(z) = z^n \)`where \( n = 1 \) is our base case. We've established in part (a) that f(z) = z is continuous, which serves as the base case for \( n = 1 \). Now, we need to show that if
`\( g(z) = z^n \) is continuous at \( z = z_0 \),` then
`\( g(z) = z^(n+1) \)`is also continuous at
`\( z = z_0 \).` This proof involves using the continuity of
`\( g(z) \)` to show the continuity of
`\( g(z) \)` raised to the power of
`\( n+1 \).`