Showing a holomorphic function f is constant under certain conditions Let $f$ be holomorphic on the open unit disk $D = \{z : \space|z| < 1 \}$. Show that if any
of the following conditions holds, then f is constant in D:
i. $f'= 0$ everywhere in D
ii. $f$ is real-valued in D
iii. $|f|$ is constant in D
iv. $\arg(f)$ is constant in D
How do you show all these?
Im thinking for the first one it would maybe be to do with the cauchy riemann equations but im not sure how to appoach this? and the I have an idea for the third that by liouvilles theorem $|f|$ is bounded and because it is holomorphic in this range it is therefore constant?
 A: Hints: For (i) and (ii), use the fact that holomorphic implies you can expand $f$ as a Taylor series in $z$.
For (iii), you cannot use Liouville's theorem because that requires $f$ to be entire (holomorphic on the whole complex plane). However, for (iii) and (iv) you might be able to do something with the fact that $\log(f) = \log(|f|) + i \arg(f)$, where $\log(f), \arg(f) $ are only defined locally on some region with an appropriate branch cut.
A: i) Let $f=u+iv.$ Then $f'= u_x+iv_x = v_y-iu_y.$ Thus $u_x,u_y,v_x,v_y$ are all $0.$ You're back in calculus: Both $u,v$ are constant, hence $f$ is constant.
ii), iii), iv) follow quickly from the open mapping theorem. (As pointed out in the comments, Liouville doesn't apply here.)
A: If $f$ is analytic on $D$, $f'(z)$ is analytic also.  We have, therefore, for any $z_1,z_2\in D$
$$f(z_2)-f(z_1)=\int_{z_1}^{z_2} f'(z)\,dz=0\implies f(z_2)=f(z_1)\implies f\,\text{is a constant}$$

If $f=u+iv$ is analytic on $D$, then $u_x=v_y$ and $u_y=-v_x$.  If $v=0$ on $D$, then $u_x=u_y=0$, which implies $u$ is a constant.

If $f=u+iv$ is analytic on $D$, then $u_x=v_y$ and $u_y=-v_x$.  If $|f|$ is constant, then $u^2+v^2$ is likewise.  Therefore, differentiating reveals 
$$uu_x+vv_x=0\tag 1$$ 
and 
$$uu_y+vv_y=0 \tag 2$$
Solving $(1)$ and $(2)$ simultaneously yields $u_x^2+u_y^2=v_x^2+v_y^2=0$, from which we see that $u$ and $v$ are constants.

If $f=u+iv$ is analytic on $D$, and $\arg(f)$ is constant on $D$ then $u=Cv$ for some constant $C$.  Using the Cauchy-Riemann Equations, we find that if $u=Cv$, then $u_x^2+u_y^2=v_x^2+v_y^2=0$, which implies $u$ and $v$ are constants.  Therefore, $f$ is a constant.
