Question

Which of the statements given below is/are false?
(I) \(f(z)=x^2+y^2+2ixy\) (\(z=x+iy\)) is differentiable only at the points that lie on the x-axis.
(II) \(f(z)=|z-1|^2\) is differentiable at \(z=1\), but not analytic at \(z=1\).
(III) There exists an analytic function in \(\mathbb{C}\) whose imaginary part is \(x^2\).

• A. Only (III)
• B. Only (I) and (II)
• C. Only (II) and (III)
• D. All (I), (II) and (III)

Hint:

Apply the Cauchy-Riemann equations to (I) and (II), and the harmonic condition to (III).

Answer 1

The Correct Option is A

Step 1: Use the Wirtinger derivative \(\partial/\partial\bar z=\frac{1}{2}(\partial/\partial x+i\,\partial/\partial y)\); a function is complex-differentiable at a point exactly where this derivative vanishes there. For (I), \(f=x^2+y^2+2ixy\) gives \(f_x=2x+2iy\), \(f_y=2y+2ix\), so \[\frac{\partial f}{\partial\bar z}=\frac{1}{2}\big[(2x+2iy)+i(2y+2ix)\big]=\frac{1}{2}\big[2x+2iy+2iy-2x\big]=2iy.\] This vanishes exactly when \(y=0\), confirming (I) is true using Wirtinger calculus instead of the standard Cauchy-Riemann pair.

Step 2: For (II), set \(w=z-1\), so \(g=w\bar w\). Treating \(w,\bar w\) as independent variables, \(\partial g/\partial\bar w=w\), which vanishes only at \(w=0\), i.e. \(z=1\). So \(g\) is differentiable only at \(z=1\) and fails to be analytic there since it is not differentiable in any full neighborhood, confirming (II) is true.

Step 3: For (III), attempt to directly construct a harmonic conjugate \(u\) for \(v=x^2\) using the Cauchy-Riemann pair \(u_x=v_y\) and \(u_y=-v_x\). Since \(v_y=0\), we need \(u_x=0\), so \(u=\phi(y)\) for some function of \(y\) alone. Since \(v_x=2x\), we also need \(u_y=-2x\), i.e. \(\phi'(y)=-2x\). But the left side depends only on \(y\) while the right side depends on \(x\), which is impossible unless both sides are simultaneously constant in \(x\) and equal to \(-2x\), a contradiction.

Step 4: No such conjugate exists, so (III) is false, and it is the only false statement. \[\boxed{\text{Only (III)}}\]