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: $\frac{a}{a^2+b^2}$