Show that holomorphic function $f: \mathbb{C} \rightarrow \mathbb{C}$ is constant Let's $f: \mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function such that values $f$ are on line $y=ax+b$. Show that $f$ is constant.
I think I should use Cauchy-Riemann equations but I don't know what does mean that values $f$ are on line $y=ax+b$. Can you explain me that?
Thanks in advance.
 A: Your hypothesis reads: $$f(x+iy) = u(x,y) + i(a u(x,y) + b),$$ that is, the point $f(x+iy)$ as a point in $\Bbb R^2$ belongs to the said line. The Cauchy-Riemann equations read, abbreviating the notation: $$u_x = a u_y, \quad u_y = -au_x.$$ So $u_x = -a^2u_x \implies (1+a^2)u_x = 0$ and... oh, you can conclude now! (if you need one more push, please tell me)
A: Herewith my second answer:
Holomorphic functions are conformal except where their derivatives are zero.  It can't be conformal if its image is a straight line.
A: I'll assume you meant $x$ and $y$ are the real and imaginary parts of $f(z)$ and $a$ and $b$ are real.
Suppose $f'(z_0)\ne 0$.  Choose $\varepsilon$ so close to $0$ that when $|\Delta z|=\varepsilon$ then $f(z_0+\Delta z)$ differs from $f(z_0)+f'(z_0)\,\Delta z$ by less than $\Delta z/1000000$.  Then as $\Delta z$ goes around the circle $\{z:|z-z_0|=\varepsilon\}$, then $f(z_0)+f'(z_0)\,\Delta z$ goes around the circle $\{w:|w-f(z_0)| = |f'(z_0)|\varepsilon\}$.  That circle is not on a straight line, and $f(z)$ stays close to $f(z_0)+f'(z_0)\,\Delta z$.
A: This also follows immediately from the open mapping theorem.
A: How about this: you can rotate/translate your line such that your new line matches with real axis (or with imaginary axis, your chose). Now, that new function, say $g=g(z)$, has $\operatorname{Im}(g)=0$, so you can use C-R equations to prove that $g=\operatorname{const}.$, and from that you can find that $f$ is also constant (because $g(z)=(f(z)-b)e^{-i\alpha}$, where $\alpha = \arctan a$, something like that).
