Prove: $f$ Is A Constant Fucntion

Let $f(z)=u(x,y)+iv(x,y)$ differentiable at any point at $\mathbb{C}$ s.t $u^2-v^2=C$ where $C$ Is constant, prove $f(z)$ is constant

$$(f(z))^2=[u(x,y)+iv(x,y)]^2=u^2-v^2+2iuv$$

So we have $f(z)^2=c+2iuv$

But what can we say about $f(z)^2$ or $2iuv$?

• Calculate $\|f(z)\|$. It would be useful – C.F.G Aug 14 '17 at 15:49
• @C.F.G can you explain elaborately why $\|f(z)\|$ is useful? – MAN-MADE Aug 14 '17 at 16:09
• I made mistake. – C.F.G Aug 14 '17 at 16:11

So $f(z)^2$ is constant. So $f$ can meet only two values (the two square roots of $f(z)^2$). Since $f$ is continuous, $f$ is constant.
• $f(z)^2=c$ so $f(z_1)=\sqrt{c}$ and $f(z_2)=-\sqrt{c}$? – gbox Aug 14 '17 at 17:04
• Since the domain of $f$ (namely, $\Bbb C$) is connected, this is not possible by continuity of $f$. (If the domain were not connected, the statement would be false). – ajotatxe Aug 14 '17 at 18:33
f differentiable in the complex setting means that f is holomorphic and hence that it satisfies the Cauchy-Riemann equations. Differentiating $u^2-v^2=c$ with respect to $x$ and to $y$ we obtain $uu_x-vv_x=0$ and $uu_y-vu_x=0$. Exploiting Cauchy-Riemann we get $uu_x+vu_y=0$ and $uu_y-vu_x=0$, which can be seen as a $2\times 2$ system in the unknowns $u_x$ and $u_y$. If the system has a unique solution we have that the determinant $u^2+v^2>0$, and $(u_x,u_y)=(0,0)$ is a solution, hence the unique solution. In this way we get u is constant and similarly that v is constant, imlying that f is constant. Otherwise the determinant is zero obtaining $u^2+v^2=0$, implying $u=v=0$, hence also f is identically zero. In both cases f is constant.