Problem
Suppose a and b are real numbers, not both 0. Find real numbers c and d such that $$\frac{1}{a + bi} = c + di$$
Solution
A standard trick when dealing with the reciprocal of a complex numbers is to multiply the numerator and the denominator by it’s complex conjugate, as this will make the denominator real. Applying this trick: $$c + di = \frac{1}{a + bi} = \frac{a - bi}{a - bi} \cdot \frac{1}{a + bi} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i$$ From comparing the real and imaginary parts of the left and right hand sides above: $$c = \frac{a}{a^2 + b^2} \land d= \frac{b}{a^2 + b^2}$$ Note that the solution lies on the same line as $\overline{z}$ and it’s magnitude is the reciprocal of the magnitude of $z = a + bi$.