$$
\begin{aligned}
&\text { Let } z_1=x+i y \\
&\left.\Rightarrow\left|z_1\right|=\sqrt{x^2+y^2}=1 \quad \quad \quad \because\left|z_1\right|=1\right] \\
&\text { Now, } z_2=\frac{z_1-1}{z_1+1}=\frac{x+i y-1}{x+i y+1} \\
&=\frac{x-1+i y}{x+1+i y}=\frac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)} \\
&=\frac{x^2-1-i x y+i y+i x y+i y+y^2}{(x+1)^2+y^2} \\
&=\frac{x^2+y^2-1+2 i y}{(x+1)^2+y^2}=\frac{1-1+2 i y}{(x+1)^2+y^2} \\
&=0+\frac{2 y i}{(x+1)^2+y^2} \quad\left[\because x^2+y^2=1\right]
\end{aligned}
$$
Hence, the real part of $z_2$ is zero.