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Chapter 7: Problem 28

Given the complex-valued function \(f(x, y)=(x-i y) /(x+i y),\) calculate \(|f(x, y)|^{2}\)

Short Answer

Expert verified
The square of the magnitude of the complex-valued function \(f(x, y) = (x-iy)/(x+iy)\) is equal to \(1\).

Step by step solution

01

Simplify the complex-valued function

To simplify the given function, multiply the numerator and denominator by the complex conjugate of the denominator: \((x-iy)(x-iy)/((x+iy)(x-iy))\) This simplification results in: \(f(x, y) = \frac{x^2 - (iy)^2}{x^2 + y^2} = \frac{x^2 + y^2}{x^2 + y^2} + i \frac{0}{x^2 + y^2} = 1\) Step 2: Find the magnitude of the function, \(|f(x, y)|\)
02

Find the magnitude of the complex-valued function

Since the simplified function is a constant (1), the magnitude of \(f(x, y)\) is equal to the magnitude of this constant: \( |f(x, y)| = |1| = 1\) Step 3: Square the magnitude to obtain \(|f(x, y)|^2\)
03

Square the magnitude

Since the magnitude of the function is 1, the square of the magnitude is simply: \(|f(x, y)|^2 = 1^2 = 1\) Hence, the square of the magnitude of the complex-valued function \(f(x, y)\) is equal to \(1\).

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