Problem 4 - Complex Analysis

Question

Prove the following inequalities. (a) For all \(z \in \mathbb{C}\) we have. $$ |\exp (z)-1| \leq \exp (|z|)-1 \leq|z| \exp (|z|) $$ (b) For all \(z \in \mathbb{C}\) with \(|z| \leq 1\) we have $$ |\exp (z)-1| \leq 2|z| $$

Step-by-Step Solution

Answer

The inequalities hold for all complex numbers as shown in steps, focusing on convergence and series expansions.

Step 1: Understand the complex exponential function

The complex exponential function can be expressed as a series:\[\exp(z) = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\]This series converges for all complex numbers \(z\). Understanding this series is foundational for both parts (a) and (b) of the problem.

Step 2: Consider the inequality for part (a)

We wish to show \(|\exp(z) - 1| \leq |z|\exp(|z|)\). Start by expanding \(\exp(z) - 1\) using its series expansion: \[\exp(z) - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\]By the triangle inequality:\[|\exp(z) - 1| \leq |z| + \frac{|z|^2}{2!} + \frac{|z|^3}{3!} + \cdots\]which can be rewritten as:\[|\exp(z) - 1| \leq \exp(|z|) - 1\]Continue to show: \(\exp(|z|) - 1 \leq |z|\exp(|z|)\). Notice that \(\exp(|z|) \geq 1\), leading to \(||z| |\exp(|z|) - 1| = |z|\exp(|z|) - |z|\), satisfying the inequality.

Step 3: Apply and prove the inequality for part (b)

For part (b), where \(|z| \leq 1\), expand \(\exp(z) - 1\) to first order terms, only considering up to the first term:\[\exp(z) - 1 = z + \text{higher order terms}\]For small\(|z|\), higher order terms become negligible. Hence, using the triangle inequality:\[|\exp(z) - 1| \approx |z| + |\text{small terms}|\]Thus:\[|\exp(z) - 1| \leq |z|(1 + |z| + \frac{|z|^2}{3!} + \cdots) \leq 2|z|\]For \(|z| \leq 1\), the inequality clearly holds: \(|\exp(z) - 1| \leq 2|z|\).

Key Concepts

Complex Exponential Function

Understanding the complex exponential function is key to tackling complex inequality proofs. This function, often denoted as \(\exp(z)\), can be expressed through its series expansion: \[ \exp(z) = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \] This series representation is valuable because it converges for all complex numbers \(z\), meaning it provides a way to fully describe \(\exp(z)\) for any complex number.

• The series is infinite, but each term becomes progressively smaller.
• The convergence of the series ensures \(\exp(z)\) is well-defined.

The series expansion helps in understanding how \(\exp(z)\) behaves, especially when \(z\) is near zero or belongs to a bounded region, such as a circle. By understanding its expansion, we can manipulate \(\exp(z)\) to derive useful inequalities.

Triangle Inequality

The triangle inequality is a crucial tool in proving inequalities involving complex numbers. It states that for any complex numbers \(a\) and \(b\), the following holds: \[ |a + b| \leq |a| + |b| \] This property is incredibly useful when dealing with series expansions of complex functions. For example, when expanding \(\exp(z) - 1\) using its series, we have: \[ \exp(z) - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \] Using the triangle inequality on this series:

• We can sum the absolute values of each term separately.
• This leads to inequalities that provide upper bounds, which help in proving the desired inequalities.

This step is analytical, creating pathways to approximate or bound \(|\exp(z) - 1|\) by simpler or more useful expressions.

Series Expansion

Series expansions are powerful mathematical tools, especially in complex analysis. They allow you to write functions as infinite sums of terms. For example, the expansion of the exponential function \(\exp(z)\) is: \[ \exp(z) = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \] This expansion is valuable because each term gives a glimpse into the behavior of \(\exp(z)\).

• The leading terms (e.g., \(1 + z\)) capture the primary behavior of the function.
• The higher order terms (e.g., \(\frac{z^2}{2!}, \frac{z^3}{3!}\)), though smaller, become essential when higher precision is needed.

For small values of \(z\) or when \(|z| \leq 1\), as stated in part (b), truncating after a couple of terms often yields a good approximation, showing why \(\exp(z) - 1 \approx z\). As such, series expansions offer a practical approach to understanding the behavior of complex functions within certain bounds.

Convergent Series

A convergent series is a series whose partial sums become closer and closer to a certain value as more terms are added. In the context of the complex exponential function, the series \[ \exp(z) = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \] converges for all complex \(z\).

• This convergence is crucial as it ensures that \(\exp(z)\) is well-behaved for all complex values of \(z\).
• The properties of convergence help in validating inequalities, especially when manipulating series using the triangle inequality.

For instance, knowing that the series converges assures us that approximations made using initial terms hold validity within specified bounds. Whether it’s bounding \(|\exp(z) - 1|\) or showing domains like \(|z| \leq 1\) do not lead to divergence, convergence guarantees the reliability of such analysis.