Discussion
If \(z = \frac{3+i}{2-i}\) then \(z^{-1}\) is equal to
Answer: C
\(z^{-1}\) is \(\frac{1-i}{2}\).
Simplify z by multiplying numerator and denominator by the conjugate \(2 + i\):
\[ \begin{aligned} z &= \frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6 + 5i + i^{2}}{5} \\ &= \frac{5 + 5i}{5} = 1 + i \end{aligned} \]
\[ \begin{aligned} z^{-1} &= \frac{1}{1+i} = \frac{1-i}{(1+i)(1-i)} \\ &= \frac{1-i}{2} \end{aligned} \]
1 + i is z itself, and \(1 - i\) is its conjugate, which equals \(z^{-1}\) only when \(|z| = 1\); here \(|z|^{2} = 2\), so the conjugate must be divided by 2.
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