Proof of the Inverse Function Theorem in Complex Analysis 📂Complex Anaylsis

Proof of the Inverse Function Theorem in Complex Analysis

Theorem ¹

If a function $f : \mathbb{C} \to \mathbb{C}$ is analytic at $\alpha$ and $f ' (\alpha) \ne 0$, then $f^{-1}$ exists on $\mathcal{N} \left( f(\alpha) \right)$.

Explanation

Let us think carefully about the condition $f ' (\alpha) \ne 0$.

If we think of it as a real function, it means the function is either increasing or decreasing, which is the condition for the existence of an inverse function. Borrowing a geometric expression, it refers to a smooth function, meaning there are no bent points such as points where the direction suddenly turns. What must be noted in the inverse function theorem is that even if this condition is satisfied, it does not mean the inverse function itself exists, but rather that there is a local limitation.

Proof

It suffices to show that for $w \in \mathcal{N} (f(\alpha))$, the equation $w = f(z)$ has a unique solution.

$$ \beta := f(\alpha) \\ g(z) := f(z) - \beta $$ If we set this, then $g(\alpha) = 0$ and $g ' (\alpha) \ne 0$. This means that $\alpha$ is a simple zero of $g$, and there exists $\rho >0$ satisfying $g(z) \ne 0$ on $|z - \alpha | \le \rho$.

For the circle $\mathscr{C}: |z - \alpha| = \rho$, let us define $h(z) := -\gamma$ satisfying $$ m := \min_{\mathscr{C}} |g(z)| \\ |\gamma| < m $$ Then the following holds on $\mathscr{C}$. $$ g(z) \ne 0 \\ |h(z) | = | - \gamma | = |\gamma| < m \le |g(z)|$$

Rouché’s theorem: If $g$ and $h$ are analytic on a simple closed contour $\mathscr{C}$ and satisfy $|h(z)| \le |g(z)|$ on $\mathscr{C}$, then $g$ and $g + h$ have the same number of zeros in the interior of $\mathscr{C}$.

By Rouché’s theorem, $g$ and $g + h = g - \gamma$ have the same number of zeros in the interior of $\mathscr{C}$.

However, as seen earlier, since $g$ had only one simple zero $\alpha$, the zero satisfying $g(z) - \gamma = 0$ is also the only one in the interior of $\mathscr{C}$. In summary, we can say that the equation $g(z) = \gamma$ has only one solution in the interior of $\mathscr{C}$.

Now, if we let $w = \beta + \gamma$, then $$ f(z) - \beta = w - \beta $$ that is, $w = f(z)$ has a unique solution in the interior of $\mathscr{C}$, $\mathcal{N}(\alpha): |z - \alpha| < \rho$.