4

Theorem :- Let $f(z) =u(x,y) +iv(x, y)$ If, $f'(z)=0$ everywhere in a domain $D$ then $f(z)$ must be constant throughout D. Here

In this theorem, by using cauchy reimann equations, we get $u_x =v_y=0$ and $u_y=-v_x=0$

Now I am not getting why it does not suffice to show that f is constant.

Please help me with this, (I am thinking that both partial derivatives are zero than $u(x, y)$ is constant similarly $v(x, y)$) I don't understand what I am missing.

1 Answers1

1

As pointed out by our OP user772744 in the text of the question itself,

$f'(z) = 0 \tag 1$

for

$z \in D \tag 2$

implies

$u_x = u_y = v_x = v_y = 0 \tag 3$

in $D$ as well; that is

$\nabla u = \nabla v = 0 \tag 4$

in $D$.

Now, since $D$ is a connected set in $\Bbb C$, it is path connected; see this wikipedia entry. And since $D$ is a path connected set in $\Bbb C$, it is $C^1$ path connected, see this question and its answers; that is, any two points

$z_1, z_2 \in D \tag 5$

may be joined by a $C^1$ path

$\gamma:I = [0,1] \to D \tag 6$

with

$\gamma(0) = z_1, \gamma(1) = z_2; \tag7$

such a path $\gamma(t)$ may of course be viewed in either of two ways, as

$\gamma:I \to \Bbb C, \tag{7.1}$,

with

$\gamma(t) = x(t) + iy(t), \tag{7.2}$

or as

$\gamma:I \to \Bbb R^2, \tag{7.3}$

with

$\gamma(t) = (x(t), y(t)); \tag{7.4}$

for any such points $z_1$, $z_2$ and such a real path $\gamma(t)$ as in (7.4) we have the real integral formula

$u(z_2) - u(z_1) = u \circ \gamma(1) - u \circ \gamma(0) = \displaystyle \int_{z_1}^{z_2} \dfrac{d(u \circ \gamma)(t)}{dt} \; dt$ $= \displaystyle \int_{z_1}^{z_2} \nabla u \cdot \gamma'(t) \; dt = \int_{z_1}^{z_2} 0 \cdot \gamma'(t) \; dt = 0, \tag 8$

where

$\gamma'(t) = (x'(t), y'(t)), \tag{8.1}$

yielding

$u(z_2) = u(z_1); \tag 9$

likewise,

$v(z_2) = v(z_1); \tag{10}$

that is, $u$ and $v$ are constant in $D$; hence, $f(z)$ is constant as well. $OE\Delta$.

Note Added in Edit, Sunday 27 December 2020, 6:35 PM PST: The above demonstration takes its cue from our OP user772744's observation that (1) implies (3); that is, it reduces the problem to integrals of the form (8) which compute $u(z_2) - u(z_1)$ and $v(z_2) - v(z_1)$ separately. It is of course possible to work directly from (1):

$f(z_2) - f(z_1) = \displaystyle \int_{z_1}^{z2} f'(z) \; dz = \int_{z_1}^{z2} 0 \;dz = 0, \tag{11}$

where now $\gamma(t)$ is of the form (7.2), so that

$\gamma'(t) = x'(t) + iy'(t), \tag{11.1}$

and

$dz = \gamma'(t) \; dt. \tag{12}$

End of Note.

Robert Lewis
  • 71,180
  • 1
    If $z,z'\in D$ and $f(z)=f(z')$ there is an open ball $B$ centered at $z'$ with $B\subset D.$ Now if $z'=x'+iy'$ then for any $z''=x''+iy''$ in $B$, the line-segments from $z'$ to $x''+iy'$ and from $x'' +iy'$ to $z''$ are in $B,$ hence in $D.$ Integrating $f'$ from $z'$ to $z''$ along these segments shows $f(z'')=f(z')=f(z).$ So $f^{-1}{f(z)}$ is open. So $F={ f^{-1}{f(z)}}: z\in D}$ is a pair-wise disjoint family of non-empty open subsets of $D$ with $\cup F=D.$ But $D$ is connected so $F={D}$ so for some (any) $z\in D$ we have $f^{-1}{f(z)}=D.$ – DanielWainfleet Dec 31 '20 at 22:37