I don't understand how it can help?
$\displaystyle e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}$
$\displaystyle -f(z) = -u(x,y) - iv(x,y)$. So $\displaystyle e^{-f(z)}=e^{-u(x,y)}(\cos v(x,y)-i\sin v(x,y))$. You are told that $\displaystyle u(x,y)\geqslant0$ for all x,y. What does that tell you about $\displaystyle |e^{-f(z)}|$ ?
$\displaystyle -f(z) = -u(x,y) - iv(x,y)$. So $\displaystyle e^{-f(z)}=e^{-u(x,y)}(\cos v(x,y)-i\sin v(x,y))$. You are told that $\displaystyle u(x,y)\geqslant0$ for all x,y. What does that tell you about $\displaystyle |e^{-f(z)}|$ ?
Thanks.
Maybe I should apply Liouville's theorem
, but I have to prove that function $\displaystyle e^{-f(z)}$- an entire, given that $\displaystyle f(z)$ - entire function...how can i prove that $\displaystyle e^{-f(z)}$ entire function also?
Thanks.
Maybe I should apply Liouville's theorem
, but I have to prove that function $\displaystyle e^{-f(z)}$- an entire, given that $\displaystyle f(z)$ - entire function...how can i prove that $\displaystyle e^{-f(z)}$ entire function also?
Yes, that is exactly what you need to do. In fact, a differentiable function of a differentiable function is always differentiable (that is basically what the chain rule says).