If |z1| = 1 (z1 ≠ –1) and z2 = (z1 – 1) / (z + 1),

According to the question,

Let z1 = x + iy

$\Rightarrow\left|z_{1}\right|=\sqrt{x^{2}+y^{2}}=1$

$\mathrm{Z}_{2}=\frac{\mathrm{z}_{1}-1}{\mathrm{z}_{1}+1}=\frac{\mathrm{x}+\mathrm{iy}-1}{\mathrm{x}+\mathrm{iy}+1}$

$=\frac{[(x-1)+i y][(x+1)-i y]}{[(x+1)+i y][(x+1)-i y]}$

$=\frac{\left(x^{2}-1\right)+y^{2}+i[(x+1) y-(x-1) y]}{(x+1)^{2}+y^{2}}$

$=\frac{x^{2}+y^{2}-1+2 i y}{(x+1)^{2}+y^{2}}$

Since $x^{2}+y^{2}=1$

$=\frac{1-1+2 i y}{(x+1)^{2}+y^{2}}$

$=0+\frac{2 i y}{(x+1)^{2}+y^{2}}$

Therefore, the real part of z2 is zero.