Random variable with pdf proportional to Normal

Question

I don't understand the step highlighted in green.

I know $f_Z(z)= \frac{k}{\sqrt{2\pi}}$ $ e^{-\frac{z^2}{2}}$ when $z>-\frac{\mu}{\sigma}$

and $0$ elsewhere;

but i'm stuck at this point.

Answer 1

The term in the denominator just normalizes the pdf of $Z$ such that

$$\int_{-\infty}^\infty f_Z(z) \, dz=\int_\kappa^\infty f_Z(z) \, dz=1$$

is satisfied.