Ta có:

$\begin{align}   & z=a+bi;\left| z \right|=1\Leftrightarrow {{a}^{2}}+{{b}^{2}}=1 \\  & \Rightarrow \frac{{{z}^{2}}-1}{z}=\frac{{{\left( a+bi \right)}^{2}}-1}{a+bi}=\frac{\left( {{a}^{2}}-{{b}^{2}}-1 \right)+2abi}{a+bi} \\ \end{align}$

$\begin{align}  & \Rightarrow \frac{{{z}^{2}}-1}{z}=\frac{\left[ \left( {{a}^{2}}-{{b}^{2}}-1 \right)+2abi \right]\left( a-bi \right)}{\left( a+bi \right)\left( a-bi \right)} \\  & \Rightarrow \frac{{{z}^{2}}-1}{z} \\ \end{align}$

$\begin{align}  & =\frac{a\left( {{a}^{2}}-{{b}^{2}}-1 \right)+2a{{b}^{2}}+\left[ 2{{a}^{2}}b-\left( {{a}^{2}}-{{b}^{2}}-1 \right) \right]i}{{{a}^{2}}+{{b}^{2}}} \\  & \Rightarrow \frac{{{z}^{2}}-1}{z}=\frac{a\left( {{a}^{2}}+{{b}^{2}}-1 \right)+b\left( {{a}^{2}}+{{b}^{2}}+1 \right)i}{{{a}^{2}}+{{b}^{2}}} \\  & \Rightarrow \frac{{{z}^{2}}-1}{z}=2bi\left( {{a}^{2}}+{{b}^{2}}=1 \right) \\ \end{align}$

Vậy đáp án đúng là B.