# If argument of $f$ is constant, then $f$ is constant

$f$ is holomorphic on an open unit disk $D$. If $arg(f)$ is constant in $D$, then $f$ is constant in $D$

I don't know where to start:

I know the Cauchy Riemann Equations (these must be satisfied for complex differentiability) : $u_x=v_y$ and $v_x=-u_y$

I know the open mapping theorem - every open set is mapped to an open set

I know the maximum modulus theorem - if $f$ is non-constant, then it can't attain a maximum in $D$.

• Constant argument means that the image lies inside a particular line, which has empty interior. Now, apply open mapping theorem. – Crostul Mar 19 '16 at 18:46
• Suppose the argument is constantly $\theta$. Then $f(z)e^{-\theta i}$ is still holomorphic, but is real on the entire $D$. Now use Cauchy-Riemann. – hmakholm left over Monica Mar 19 '16 at 18:46
• Thanks, so if a function is real, I can rewrite the function as $u(x,y)+iv(x,y), but since the imaginary part is$0$, I need to somehow show that the derivatives are both$0$(using C-R), thus making$f'(x)=0$, implying that it – GRS Mar 19 '16 at 18:58 • Thanks, so if a function is real, I can rewrite the function as$u(x,y)+iv(x,y)$,but since the imaginary part is$v(x,y)=0$,but since the imaginary part is$0$, I need to some how show that the derivatives are both$0$, implying that$v_y=u_x$, and I can show that$u'(x,y)=0$– GRS Mar 19 '16 at 19:03 ## 1 Answer If the argument of$f$is constant, writing$f$as:$f = |f| \exp(\arg f)$, we see that$|f|$must be holomorphic. Since$|f|\$ has a constant imaginary part, it must be itself constant.

• So does the open mapping theorem also work for: Constant get mapped to constant, closed get mapped to closed? – GRS Mar 19 '16 at 19:07