Understanding complex numbers and their absolute values is an essential topic in mathematics, particularly in advanced algebra and complex analysis. In this article, we will delve into the intriguing question, “Which complex number has an absolute value of 5?” We will explore the principle of absolute value in the context of complex numbers, derive the related mathematical expressions, and provide examples to facilitate comprehension.
Complex numbers are expressed in the form of a + bi, where a is the real part, b is the imaginary part, and i represents the imaginary unit defined by the property that i^2 = -1. The absolute value, or modulus, of a complex number provides insight into its distance from the origin in the complex plane. This distance is calculated via the formula:
|z| = √(a² + b²)
Here, Next, squaring both sides of the equation eliminates the square root:
This equation defines a circle with a radius of 5, centered at the origin (0,0) in the complex plane. Consequently, any complex number To gain a comprehensive understanding of the complex numbers that satisfy this condition, we can express This formulation elucidates how varying values of To illustrate, let us consider some specific values for Thus, one complex number is This results in the complex number Thus, we can identify the complex number Here, the complex number simplifies to However, we must also consider the negative scenarios. Each value of This leads to the complex number This encompasses the complex number In summary, a plethora of complex numbers can claim an absolute value of 5. These include:
In closing, complex numbers that adhere to the condition of having an absolute value of 5 are represented mathematically by the equation denotes the absolute value, while a and b are derived from the complex number z = a + bi. To determine which complex numbers exhibit an absolute value of 5, we establish the equation:
√(a² + b²) = 5a² + b² = 25z = a + bi that falls on this circle will possess an absolute value of 5.
b in terms of a:
b = √(25 - a²)a result in corresponding values for b. By substituting values for a, we will derive various complex numbers.
a:
a = 0: b = √(25 - 0²) = √25 = 50 + 5i or simply 5i.
a = 3: b = √(25 - 3²) = √(25 - 9) = √16 = 43 + 4i.
a = 4: b = √(25 - 4²) = √(25 - 16) = √9 = 34 + 3i.
a = 5: b = √(25 - 5²) = √(25 - 25) = √0 = 05 + 0i or just 5.
a yields a corresponding negative value for b, thus producing additional complex numbers with absolute values of 5. For instance:
a = 3: b = -√(25 - 3²) = -43 - 4i.
a = 4: b = -√(25 - 4²) = -34 - 3i.
0 + 5i3 + 4i4 + 3i5 + 0i3 - 4i4 - 3i0 - 5ia and b.a² + b² = 25. The vast array of solutions demonstrates the infinite nature of such complex numbers, providing students with an enthralling glimpse into the world of complex analysis, geometry, and algebra alike. Understanding these principles not only enriches one’s mathematical acumen but also equips learners with tools for navigating more intricate mathematical concepts and applications in the future.
