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Problem 21

# Suppose $$f(z)$$ is analytic at $$z_{0}$$ with $f^{\prime}\left(z_{0}\right) \neq 0 .$$Show that there exists an analytic function$$g(z)$$such that$$f(g(z))=z$ in some neighborhood of $$z_{0}$$. This is known as the inverse function theorem.

Expert verified
By applying Newton's method and carefully controlling the error of each iteration, we can show that there exists an analytic function $$g$$ whose composition with $$f$$ equals $$z$$ in some neighborhood of $$z_{0}$$.
See the step by step solution

## Step 1: Setup the Problem

We want to find a function $$g(z)$$ such that $$f(g(z)) = z$$, and we know that $$f$$ is analytic and non-derogatory at $$z_{0}$$. Initialize $$g$$ as $$g_{0}(z) = z_{0} + (z - f(z_{0}))/f'(z_{0})$$. Now, for any $$z$$ in a neighborhood of $$z_{0}$$, we replace $$g_{0}(z)$$ with $$z$$ and try to solve for $$g_{0}(z)$$. This can be done due to the fact that $$f'$$ is continuous and not zero in a neighborhood of $$z_{0}$$.

## Step 2: Apply Newton's Method

Applying Newton's method in the complex domain for $$f(g(z)) - z = 0$$ we get: $$g_{n+1}(z) = g_{n}(z) - (f(g_{n}(z)) - z)/(f'(g_{n}(z))),$$. Now, this series of $$g_n(z)$$ needs to be shown to converge to a function $$g(z)$$ that satisfies $$f(g(z))=z$$.

## Step 3: Prove Convergence

Let's denote $$e_{n}(z) = f(g_{n}(z)) - z,$$. The error after the $$n+1$$ application of Newton's method is given by $$e_{n+1}(w) = f(w) - f(g_{n}(z)) + e_{n}(z) (f'(w)-f'(g_{n}(z))) / f'(g_{n}(z))$$. Since $$f$$ is analytic at $$z_{0}$$ and $$f'(z_{0}) ≠ 0$$, $$f(w)$$ and $$f'(w)$$ can be written as power series centered at $$w = g_{n}(z)$$. Therefore, we see that $$|e_{n+1}(z)| < |e_{n}(z)|$$. This shows that the sequence of $$e_{n}(z)$$ approaches 0 and hence $$g_n(z)$$ converges to some function $$g(z)$$ in the neighborhood of $$z_{0}$$.

## Step 4: Verify $$f(g(z)) = z$$

$$(f \circ g)(z) = f(g_{n}(z)) + e_{n}(z)$$. As $$n$$ tends to infinity, $$e_{n}(z)$$ tends to zero, and hence $$(f \circ g)(z)$$ tends to $$z$$ in the neighborhood of $$z_{0}$$. This confirms that $$f(g(z)) = z$$ in the neighborhood of $$z_{0}$$.

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