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Real Analysis Theorem: For any pair of real numbers a and b with aLet a and b be a pair of real numbers with a

User GeekzSG
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Final answer:

The real analysis theorem in question shows that multiplying a complex number by its conjugate results in a real number, as the formula A*A* = (a + ib)(a - ib) = a^2 + b^2 demonstrates, by canceling out the imaginary parts.

Step-by-step explanation:

The question revolves around a concept in real analysis, specifically related to complex numbers and their properties. When we have a complex number A given by a + ib, where a and b are real constants and i represents the imaginary unit (i.e., the square root of -1), we can find the product of A and its complex conjugate A*. The complex conjugate of A is a - ib. When we multiply A by A*, the result is A * A* = (a + ib) (a − ib) = a² + b², which is a real number as the imaginary parts cancel out. This property is useful in various applications because it can turn a complex equation into a real one, making it easier to solve or analyze.

User Antono
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