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: For (I), write \(f=u+iv\) with \(u=x^2+y^2\), \(v=2xy\). The Cauchy-Riemann equations require \(u_x=v_y\) and \(u_y=-v_x\). Here \(u_x=2x=v_y=2x\) holds always, while \(u_y=2y\) and \(-v_x=-2y\) requires \(2y=-2y\), i.e. \(y=0\). Since \(u,v\) have continuous partial derivatives everywhere, \(f\) is differentiable exactly where the Cauchy-Riemann equations hold, which is exactly the x-axis. This matches statement (I) exactly, so (I) is true, not false.

Step 2: For (II), let \(g(z)=|z-1|^2=(x-1)^2+y^2\), so \(u=(x-1)^2+y^2\), \(v=0\). Then \(u_x=2(x-1)=v_y=0\) requires \(x=1\), and \(u_y=2y=-v_x=0\) requires \(y=0\). So \(g\) is differentiable only at the single point \(z=1\) and nowhere else nearby, hence differentiable at \(z=1\) but not analytic there (analyticity needs differentiability throughout a neighborhood). This matches statement (II) exactly, so (II) is true, not false.

Step 3: For (III), if an analytic function \(u+iv\) had \(v=x^2\), then \(v\) would have to be harmonic, i.e. \(v_{xx}+v_{yy}=0\). Here \(v_{xx}=2\) and \(v_{yy}=0\), so \(v_{xx}+v_{yy}=2\neq0\). Since \(x^2\) is not harmonic, no analytic function can have it as its imaginary part, so statement (III) is false.

Step 4: Only (III) is false. \[\boxed{\text{Only (III)}}\]