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Introduction When we talk about "unbounded sets" in mathematics, we often describe them as sets that extend infinitely far in at least one direction. But what does that really mean? In this blog post, we'll explore a fundamental result: a set is unbounded if and only if every neighborhood of the "point at infinity" contains at least one point from that set. By the end of this post, you'll understand: What it means for a set to be bounded or unbounded. The concept of compact sets and neighborhoods of the point at infinity. A rigorous proof of our main result with intuitive explanations. This discussion assumes a basic understanding of set notation, limits, and real numbers. No knowledge of complex analysis is required! What Does It Mean for a Set to Be Bounded? A set S S in R n \mathbb{R}^n (such as the number line R \mathbb{R} or the plane R 2 \mathbb{R}^2 ) is bounded if there exists a real number M > 0 M > 0 such that every point x x ...

Introduction Have you ever wondered how expressions like ( z 1 + z 2 ) n (z_1 + z_2)^n expand so neatly into a sum of terms? Today, we’re diving into the Binomial Theorem , a powerful tool in algebra that reveals the secret behind these expansions. Our goal: to prove the formula ( z 1 + z 2 ) n = ∑ k = 0 n ( n k ) z 1 n − k z 2 k , for  n = 1 , 2 , … (z_1 + z_2)^n = \sum_{k=0}^n \binom{n}{k} z_1^{n-k} z_2^k, \hspace{3mm} \text{for } n = 1, 2, \ldots where ( n k ) = n ! k ! ( n − k ) ! , k = 0 , 1 , 2 , … , n . \binom{n}{k} = \frac{n!}{k!(n-k)!}, \hspace{3mm} k = 0, 1, 2, \ldots, n. Don’t worry if the notation seems overwhelming—we’ll break everything down step by step! What Is the Binomial Theorem? The Binomial Theorem gives a formula to expand expressions of the form ( z 1 + z 2 ) n (z_1 + z_2)^n into a sum of terms involving powers of z 1 z_1 and z 2 z_2 . Each term in the expansion involves a binomial coefficient , ( n k ) \binom{n}{k} , which tells us how many ways we c...

Introduction Ever wondered how math unlocks the mysteries of the universe? Here’s a story of how we solve a real-valued integral using complex numbers—a topic that might sound intimidating but is truly magical. Along the way, you’ll learn about the complex plane , Laurent series , poles , and the powerful Residue Theorem . Our goal is to evaluate the integral: I = ∫ 0 1 ln ⁡ ( 1 + x ) 1 + x 2   d x . I = \int_0^1 \frac{\ln(1+x)}{1+x^2} \, dx. At first glance, this looks like a tough problem. The natural logarithm ln ⁡ ( 1 + x ) \ln(1+x) combined with the 1 + x 2 1+x^2 in the denominator makes it tricky to handle directly. But with some clever thinking and the power of complex numbers, we’ll get to the solution step by step. Step 1: From Real Numbers to the Complex Plane The first step is to extend this real integral into the complex plane. This means treating x x as a complex variable z z . Our function becomes: f ( z ) = ln ⁡ ( 1 + z ) 1 + z 2 . f(z) = \frac{\ln(1+z)}{1+z^2...

Complex analysis offers elegant ways to represent and manipulate functions, and one such powerful representation is the Laurent series , which can describe functions with singularities. In this blog post, we'll explore how the exponential function exp ⁡ [ z 2 ( w − 1 w ) ] \exp\left[\frac{z}{2}\left(w - \frac{1}{w}\right)\right] can be expanded as a Laurent series, connecting it to the Bessel functions J n ( z ) J_n(z) . Laurent Series and Contour Integrals To start, let’s recall the Laurent series: If f ( z ) f(z) is analytic on an annulus R 1 < ∣ z ∣ < R 2 R_1 < |z| < R_2 , then f ( z ) f(z) can be expressed as: f ( z ) = ∑ n = − ∞ ∞ c n z n , f(z) = \sum_{n=-\infty}^\infty c_n z^n, where the coefficients are given by: c n = 1 2 π i ∫ C f ( z ) z n + 1 d z . c_n = \frac{1}{2\pi i} \int_C \frac{f(z)}{z^{n+1}} dz. Here, C C is a closed contour within the annulus. Our goal is to apply this formula to expand the exponential function as a Laurent series aro...

In this blog entry, we’ll explore a fascinating result from complex analysis, similar to the one we discussed previously about the real part of an entire function. This time, we’ll focus on the imaginary part and show that if it is bounded, the function must be constant. Here’s the statement we aim to prove: If f ( z ) f(z) is an entire function on the complex plane, and Im ⁡ ( f ( z ) ) \operatorname{Im}(f(z)) (the imaginary part of f ( z ) f(z) ) is bounded above, then f ( z ) f(z) must be constant. As before, we’ll break the proof into manageable steps and reference key theorems and ideas to make it easy to follow, even for those with just a basic background in complex analysis. Step 1: Understanding the Problem An entire function is a function that is holomorphic (analytic) everywhere on the complex plane. For f ( z ) f(z) , we’re given that its imaginary part Im ⁡ ( f ( z ) ) \operatorname{Im}(f(z)) is bounded above, meaning there exists a real number M M such that: ...