$\begin{aligned} & \text {Since, } Z=1+\cos \theta-i \sin \theta \\ & \Rightarrow z-1=\cos \theta-i \sin \theta \\ & \Rightarrow|z-1|=\left|e^{i \theta}\right|=1 \\ & \text { Now, }|z|=\sqrt{(1+\cos \theta)^2+\sin ^2 \theta}=\sqrt{2+2 \cos \theta} \\ & \therefore\left[|z-1|^2-\frac{|z|^2}{4}\right]^{\frac{1}{2}}=\left[1-\frac{1+\cos \theta}{2}\right]^{\frac{1}{2}}=\left[\frac{1-\cos \theta}{2}\right]^{\frac{1}{2}} \\ & \Rightarrow\left[|z-1|^2-\frac{|z|^2}{4}\right]^{\frac{1}{2}}=\left[\frac{1-1+2 \sin ^2 \frac{\theta}{2}}{2}\right]^{1 / 2}=\sin \frac{\theta}{2}\end{aligned}$