Liouville's Theorem in Complex Analysis

Liouville's Theorem

Art-15. Liouville's Theorem

This theorem is very important, if you read this theorem carefully then some of the theorems coming up are based on this on which you can apply it and there are many questions where you can apply this theorem and you can solve those questions with Liouville's Theorem.

Statement: If a function \( f(z) \) is analytic for all finite values of \( z \) and is bounded, then \( f(z) \) is constant.

Proof:

Let \( z_1, z_2 \) be any two points of the \( z \)-plane. Draw a circle \( C \) with center at the origin and radius \( R \), enclosing the points \( z_1 \) and \( z_2 \) so that:

\( |z_1| < R \) and \( |z_2| < R \)

Since \( f(z) \) is bounded, there exists \( M > 0 \) such that \( |f(z)| \leq M \) on \( C \).

By Cauchy's integral formula, we have

\( f(z_1) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_1} \, dz \)

and

\( f(z_2) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_2} \, dz \)

Therefore,

\[ \begin{aligned} |f(z_1) - f(z_2)| &= \left| \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_1} \, dz - \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_2} \, dz \right| \\ &= \left| \frac{1}{2\pi i} \int_C \left( \frac{1}{z - z_1} - \frac{1}{z - z_2} \right) f(z) \, dz \right| \\ &= \left| \frac{1}{2\pi i} \int_C \frac{z_1 - z_2}{(z - z_1)(z - z_2)} f(z) \, dz \right| \\ &\leq \frac{1}{2\pi i} \int_C \left| \frac{z_1 - z_2}{(z - z_1)(z - z_2)} f(z) \right| |dz| \end{aligned} \]


\[ \leq \frac{1}{2\pi} |z_1 - z_2| M \int_C \frac{|dz|}{(|z| - |z_1|)(|z| - |z_2|)} \]

\[ = \frac{1}{2\pi} \frac{|z_1 - z_2| M}{(R - |z_1|)(R - |z_2|)} \int_C |dz| \]

\[ = \frac{1}{2\pi} \frac{|z_1 - z_2| M}{(R - |z_1|)(R - |z_2|)} 2\pi R \to 0 \text{ as } R \to \infty \]

Hence \( f(z_1) = f(z_2) \)

Since this result holds for all values of \( z_1 \) and \( z_2 \) of the \( z \)-plane,

\(\therefore f(z) \) is a constant function.


There are many questions on our website, you can see them, you will find them very helpful and you must also check our YouTube channel, we have solved many questions and many theorems, which will be very helpful for you.

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