let f be a nonconstant analytic function in the domain D, show that the function $g(z)= \overline {f(z)}$ is not analytic in D

Question

let f be a nonconstant analytic function in the domain D, show that the function $g(z)= \overline {f(z)}$ is not analytic in D

---

Cauchy-Reimann equations

suppose that $$f(z)=f(x+iy)=u(x,y)+iv(x,y)$$ is differentiable at the point $z_0=x_0+iy_0$

then the partial derivatives of u and v exists and $$ \begin{aligned} f'(z_0)&=u_x(x_0.y_0)+iv_x(x_0,y_0) \\ f'(z_0)&= v_y(x_0.y_0)-iu_y(x_0.y_0) \end{aligned}$$

---

Attempt

f is a nonconstant analytic function in the domain D that means that it is diff on all D (or that $\exists \epsilon$ openball where f is diff with D)

$ f(z)=u(x,y)+iv(x,y)$ so cauchy-reimann equations hold that is $$\begin{aligned} f'(z)&=u_x(x,y)+iv_x(x,y) \\f'(z)&=v_y(x,y)-iu_y(x,y) \end{aligned} $$ we have $u_x=v_y$ and $v_x=-u_y$

Now if for g the cauchy reimman equations do not hold then g is not diff so let us check note that $g(z)=u(x,y)+-v(x,y)$ $$\begin{aligned} g'(z)&=u_x(x,y)-iv_x(x,y) \\g'(z)&=-v_y(x,y)+iu_y(x,y) \end{aligned} $$
missing step

where g is non constant

(kind of confused because i want to say that one u or v has a atleast one variable x or y. should I divide it up to some diff cases???)

and $u_x=-v_y$ and $v_x=u_y$ but that is not the case bc really $u_x=v_y$ and $v_x=-u_y$ . so the cauchy-reimann equations do not hold for $g$ then g is not diff then g is not analytic on D

Answer 1

Cauchy Rieman : $u_x=v_y, u_y=-v_x$, write $\overline f(x)=a+ib$ Suppose $\bar f$ analytic you have $a=u,b=-v$, $a_x=u_x=v_y=-b_y$ but $a_x=b_y$ since $\bar f$ is analytic, you deduce that $b_y=-b_y$. This implies that $b_y=0$. You also have $a_y=-v_x=b_x$ Since $\bar f$ is analytic $a_y=-b_x=b_x$. This implies $b_x=0$. You deduce that $a_x=a_y=b_x=b_y=0$. This implies that $f$ is constant.

Answer 2

As per the comment of one of the user (zhw), I would like to post it as an answer.

Since given that $f(z)=u+iv$ is analytic and assuming $\overline{f(z)}=u-iv$ to be analytic, their sum function $F(z)=2u$, must be an analytic function which is real and non-constant (a contradiction).