Why does $|e^{-t}t^{s-1}| = e^{-t}t^{\sigma-1}$, where $\sigma = Re(s)$?

Question

I am going over Stein and Shakarchi's section on the gamma function, and most of it makes sense except for this little technicality. I've also read a little bit of Titchmarsh and he makes an identical claim, but I still don't know why this is true.

Answer 1

Because, if we set $s=\sigma+i\tau$, $\sigma,\tau\in \bf R$, we have $$\bigl|t^{s-1}\bigr|=\bigl|t^{\sigma-1+i\tau}\bigr|=\bigl|t^{\sigma-1}\bigr|\cdot\bigl|t^{i\tau}\bigr|$$ and $t^{i\tau}$ has modulus $1$ since it is $$t^{i\tau}=\rm e^{i\tau\ln t}.$$

Answer 2

$|e^{-t}t^{s-1}| = |e^{-t}e^{\ln(t)(\sigma + j\omega - 1)}| =|e^{-t} t^{\sigma - 1}e^{j\omega \ln(t)}| = e^{-t}t^{\sigma-1}$, where $\sigma = Re(s)$