Problem 12 - Complex Analysis

Question

For \(k \in \mathbb{N}_{0}\), and \(z \in \mathbb{C}\) with \(|z|<1\), show $$ \frac{1}{(1-z)^{k+1}}=\sum_{n=0}^{\infty}\left(\begin{array}{c} n+k \\ k \end{array}\right) z^{n}=\sum_{n=0}^{\infty}\left(\begin{array}{c} n+k \\ n \end{array}\right) z^{n} $$

Step-by-Step Solution

Answer

Both series representations are valid due to the symmetry of binomial coefficients: \(\binom{n+k}{n} = \binom{n+k}{k}\).

Step 1: Understanding the Binomial Series

The expression \( \frac{1}{(1-z)^{k+1}} \) can be expanded using the generalized binomial series, which is valid for \(|z| < 1\). The binomial theorem states that for any real number \(\alpha\), \((1-x)^{-\alpha}\) can be expressed as an infinite series: \((1-x)^{-\alpha} = \sum_{n=0}^{\infty} \binom{n+\alpha-1}{n} x^n\). In our problem, \( \alpha = k+1\) and \( x = z \).

Step 2: Expansion of the Given Expression

Substitute \( \alpha = k+1 \) into the binomial series formula: \((1-z)^{-(k+1)} = \sum_{n=0}^{\infty} \binom{n+k}{n} z^n\). Each term \( \binom{n+k}{n} \) represents the binomial coefficient \( \frac{(n+k)!}{n!k!} \). Note that this coefficient is also \( \binom{n+k}{k} \), which is the number of combinations of \( n+k \) things taken \( k \) at a time.

Step 3: Confirming the Interchangeability of Binomial Coefficients

The binomial coefficients \( \binom{n+k}{n} \) and \( \binom{n+k}{k} \) are equivalent because of the symmetry property of binomial coefficients: \( \binom{n+k}{n} = \binom{n+k}{k} \). This symmetry arises from the fact that choosing \( n \) items out of \( n+k \) is the same as choosing the \( k \) items that are left.

Step 4: Conclusion of the Binomial Identity

Substituting the symmetry property, we confirm that both series representations are valid: \( \frac{1}{(1-z)^{k+1}} = \sum_{n=0}^{\infty} \binom{n+k}{k} z^n = \sum_{n=0}^{\infty} \binom{n+k}{n} z^n \). This shows the identity holds true for \(|z| < 1\) and \(k \in \mathbb{N}_0\).

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of a complex variable. The complex plane, denoted by \(\mathbb{C}\), is formed by extending the real number line to include complex numbers. Each point in the complex plane represents a complex number \(z = x + iy\), where \(x\) and \(y\) are real numbers and \(i\) is the imaginary unit.

• The modulus \(|z|\) of a complex number \(z\) is the distance from the origin \((0, 0)\) in the complex plane, defined as \(|z| = \sqrt{x^2 + y^2}\).
• Analytic functions, or holomorphic functions, in complex analysis are differentiable at every point in their domain.
• Complex analysis often involves convergence, which is crucial for assessing when infinite series like the binomial series converge for certain regions in the complex plane.

In solving problems using complex analysis, such as the binomial series expansion for \(|z| < 1\), it's important to consider how complex numbers influence convergence and the behavior of the function close to essential singular points.

Binomial Theorem

The Binomial Theorem provides a powerful way to expand expressions of the form \((a+b)^n\). It generalizes to real or complex exponents via the Binomial Series Expansion for expressions like \((1-x)^{-\alpha}\).

• For a positive integer exponent \(n\), the Binomial Theorem states \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\), where \(\binom{n}{k}\) is a binomial coefficient \(\frac{n!}{k!(n-k)!}\).
• For non-integer and complex exponents \(\alpha\), the expression \((1-x)^{-\alpha}\) can be expanded into an infinite series: \((1-x)^{-\alpha} = \sum_{n=0}^{\infty} \binom{n+\alpha-1}{n} x^n\).

In the given problem, the theorem helps us express \(\frac{1}{(1-z)^{k+1}}\) as an infinite series that uses complex analysis to handle the convergence criteria \(|z| < 1\). This expansion is essential in both theoretical and applied complex analysis.

Infinite Series

Infinite series represent the sum of infinitely many terms. Convergence of such series is a fundamental aspect of analysis.

• An infinite series \(\sum_{n=0}^{\infty} a_n\) converges if the sequence of partial sums \(s_N = \sum_{n=0}^{N} a_n\) approaches a finite limit as \(N\) goes to infinity.
• For power series \(\sum_{n=0}^{\infty} c_n x^n\), convergence depends on the radius of convergence \(R\), meaning the series converges if \(|x| < R\).

In the context of the original exercise, the convergence of the series \(\sum_{n=0}^{\infty} \binom{n+k}{k} z^n\) is guaranteed for \(|z| < 1\). This radius of convergence is crucial because going outside it can result in divergence, causing the series to not represent the function accurately.

Symmetry Property of Binomial Coefficients

Binomial coefficients, \(\binom{n}{k}\), have a fascinating property known as symmetry. Specifically, they satisfy \(\binom{n}{k} = \binom{n}{n-k}\).

• This symmetry implies that choosing \(k\) items from \(n\) is the same as leaving out \(n-k\) items, resulting in the same number of combinations.
• In the binomial series, this property enables each term to be written in more than one way, as illustrated in the original exercise, where \(\binom{n+k}{k} = \binom{n+k}{n}\).

This property is extremely useful for simplifying expressions and showing equivalencies such as in our binomial series expansion. The result that \(\sum_{n=0}^{\infty} \binom{n+k}{k} z^n = \sum_{n=0}^{\infty} \binom{n+k}{n} z^n\) uses this symmetry, affirming the identity holds true.