Let \( z = x + iy \). Then,
\[
|z - 1| = \sqrt{(x - 1)^2 + y^2}, \quad |i(z + 1)| = |i(x + 1 + iy)| = \sqrt{(x + 1)^2 + y^2}
\]
Equating:
\[
(x - 1)^2 + y^2 = (x + 1)^2 + y^2 \Rightarrow x^2 - 2x + 1 = x^2 + 2x + 1 \Rightarrow -2x = 2x \Rightarrow x = 0
\]
So, the locus is \( x = 0 \), i.e., the Y-axis.