If \( |u|=|v|=1, u v \neq-1 \), and \( z=\frac{u-v}{1+u v} \), then...
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If \( |u|=|v|=1, u v \neq-1 \), and \( z=\frac{u-v}{1+u v} \), then
(a) \( |z|=1 \)
(b) \( \operatorname{Re}(z)=0 \)
(c) \( \operatorname{Im}(z)=0 \)
(d) \( \operatorname{Re}(z)=\operatorname{Im}(z) \)
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