Calculating the Real Part of \( \frac{1}{z} \) for \( z = a + bi \)

\( \text{Re} \left( \frac{1}{z} \right) \)	where \( z = a + bi \).
= \( \frac{1}{a + bi} \)	Start with the expression for \( \frac{1}{z} \).
= \( \frac{1}{a + bi} \cdot \frac{a – bi}{a – bi} \)	Multiply by the conjugate of the denominator to rationalize it.
= \( \frac{a – bi}{a^2 + b^2} \)	Simplify the denominator using \( a^2 – (bi)^2 = a^2 – (-b^2) = a^2 + b^2 \).
= \( \frac{a}{a^2 + b^2} – \frac{b}{a^2 + b^2} i \)	Separate the fraction into real and imaginary parts.
\( \text{Re} \left( \frac{1}{z} \right) = \text{Re} \left( \frac{a}{a^2 + b^2} \right) \)	Isolate the real part of the fraction.
= \( \frac{a}{a^2 + b^2} \)	This is the real part of \( \frac{1}{z} \).

Through this process, we understand how the complex number’s conjugate is used to find the real part of its reciprocal, demonstrating an important algebraic technique in complex analysis.

Final Answer: \( \frac{a}{a^2 + b^2} \)