Consider the function \[ f(x+iy)=\sqrt{|x||y|}, \qquad x,y\in\mathbb R. \] Show that \(f\) satisfies the Cauchy-Riemann equations at the origin, yet \(f\) is not holomorphic at \(0\).
Solution
1. Write \(f=u+iv\), where\[ u(x,y)=\sqrt{|x||y|}, \qquad v(x,y)=0. \]
The function \(f\) is real-valued, so its imaginary part is identically zero.
2. All four first partial derivatives \(u_x(0,0)\), \(u_y(0,0)\), \(v_x(0,0)\), and \(v_y(0,0)\) are zero.
Because \(u(h,0)=u(0,h)=0\) for every real \(h\), one has \[ u_x(0,0)=\lim_{h\to0}\frac{u(h,0)-u(0,0)}h=0, \] and similarly \(u_y(0,0)=0\). Since \(v\equiv0\), also \(v_x(0,0)=v_y(0,0)=0\).
3. Hence the Cauchy-Riemann equations hold at the origin.
At \((0,0)\), \[ u_x=v_y=0, \qquad u_y=-v_x=0, \] by 2.
4. Along the real axis, the difference quotient of \(f\) at \(0\) is zero.
For real \(t\ne0\), \[ \frac{f(t)-f(0)}t=\frac{0}{t}=0, \] because \(f(t)=u(t,0)=0\).
5. Along the line \(z=t(1+i)\) with \(t>0\), the difference quotient is \(1/(1+i)\).
For \(t>0\), \[ f(t+it)=\sqrt{|t||t|}=t, \] so \[ \frac{f(t(1+i))-f(0)}{t(1+i)}=\frac1{1+i}. \]
6. Therefore \(f\) is not complex differentiable at \(0\), and hence is not holomorphic at \(0\).
The limits in 4 and 5 approach different values as \(t\to0^+\), so the complex difference quotient has no limit at the origin.