|e^ix + 1| = \sqrt(\cos x + 1)^2 + \sin^2 x = \sqrt\cos^2 x + 2\cos x + 1 + \sin^2 x = \sqrt2 + 2\cos x - Sterling Industries

Title: Understanding |e^{ix} + 1|: A Deep Dive into Complex Exponential and Trigonometric Identities

📅 April 18, 2026 👤 scraface

Apr 18, 2026

Title: Understanding |e^{ix} + 1|: A Deep Dive into Complex Exponential and Trigonometric Identities

|e^{ix} + 1| = √(2 + 2cos x): Unlocking the Beauty of Complex Numbers and Trigonometry

Understanding the Context

The expression |e^{ix} + 1| may appear abstract at first glance, but behind this elegant formula lies a powerful connection between complex analysis, trigonometry, and real-world applications. In this article, we’ll explore how this modulus evaluates to √(2 + 2cos x), uncover its geometric and algebraic interpretations, and highlight its relevance in engineering, physics, and mathematics.

What Does |e^{ix} + 1| Represent?

The symbol e^{ix} is a complex exponential rooted in Euler’s formula:

$|e^{ix} + 1| = \sqrt{(\cos x + 1)^2 + \sin^2 x} = \sqrt{\cos^2 x + 2\cos x + 1 + \sin^2 x} = \sqrt{2 + 2\cos x}$

Key Insights

> e^{ix} = cos x + i sin x

Adding 1 gives:

> e^{ix} + 1 = (1 + cos x) + i sin x

The modulus (| |) of a complex number a + bi is defined as:

> |a + bi| = √(a² + b²)

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Final Thoughts

Applying this:

> |e^{ix} + 1| = √[(1 + cos x)² + (sin x)²]

Expanding (1 + cos x)²:

= √[1 + 2cos x + cos² x + sin² x]

Using the fundamental Pythagorean identity:

> cos² x + sin² x = 1

we simplify:

= √[1 + 2cos x + 1] = √(2 + 2cos x)

Hence:

> |e^{ix} + 1| = √(2 + 2cos x)