# Function analytic on a disk maps linear segments to line segments

Suppose f is analytic on a disk D and the image of every horizontal line segment is a horizontal line segment.

Can I get some suggestions for an approach to proving that the derivative of f is real-valued?

Thanks.

Using the suggestions below, I now have the following:

Let $\gamma(t)$ be a parametrization of a horizontal segment. Then:

$$\frac{d}{dt}f(\gamma(t))=f'(\gamma(t))\gamma'(t)$$

This means that the tangent to the path $f(\gamma(t))$ is found by multiplying the tangent $\gamma'(t)$ to the path of $\gamma(t)$ by $f'(\gamma(t))$. In order for the path $f(\gamma(t))$ to remain a horizontal segment, the argument of $f'(z)$ must equal zero.

Now, by the Cauchy-Riemann conditions, if $f=u+iv$, then $$f'=u_x+iv_x$$ and $$f'=v_x-iu_y.$$

Thus, in order that the argument of $f'(z)$ equals zero, I must have both $$v_x=0\quad\text{and}\quad u_y=0.$$

Note that $$v_x=0\implies v=\phi(y),$$ that is, $v$ is strictly a function of $y$.

Similarly, note that $$u_y=0\implies u=\psi(x),$$ that is, $u$ is strictly a function of $x$.

However, the Cauchy-Riemann condition $$u_x=v_y\implies \psi'(x)=\phi'(y)$$ for all $x$ and $y$. In my opinion, that can only happen if $$\psi'(x)=b\quad\text{and}\quad\phi'(y)=b,$$ for all $x$ and $y$, where $b$ is a real constant. Integrating, this means that $$\psi(x)=bx+a_1\quad\text{and}\quad \phi(y)=by+a_2,$$ where $a_1$ and $a_2$ are real constants. Hence, $$u=bx+a_1\quad\text{and}\quad v=by+a_2,$$ which makes $$f=u+iv=(bx+a_1)+i(by+a_2).$$ Rearranging, we get $$f=(a_1+a_2)+b(x+iy),$$ which is equivalent to saying that $$f(z)=a+bz,$$ where $a$ is a complex constant and $b$ is a real constant.

What say folks to these ideas?

Thanks.

since it is analytic so the complex derivative computed along x-direction and y-direction is the same and is the same as the derivative. you compute the derivative in the x-direction. it is real. so that is the derivative.

try to find all the functions which satisfy such conditions.

If $\gamma(t)$ is (a parametrization of) a curve and $f$ is analytic, then $\frac{d}{dt} f(\gamma(t)) = f'(\gamma(t))\gamma'(t)$ by the chain rule.

If you interpret this equality geometrically, it means that $f$ maps the tangent of $\gamma$ into $f'(\gamma(t))$ times the tangent of $f\circ \gamma$. Multiplication by the complex number $w = f'(\gamma(t))$ means multiplying the length of the tangent vector by $|w|$ and rotating it by $\arg w$.

In your problem horizontal lines map to horizontal lines, which means that $\arg f'(z) = 0$ for all $z$, or in other words that $f'$ is real-valued.

• Should this be: $$\frac{d}{dt}f(\gamma(t))=f'(\gamma(t))\gamma'(t)$$ Mar 10 '13 at 19:41
• @David, yes thank you for pointing out the typo
– mrf
Mar 10 '13 at 20:01