Question

Consider the following statements:
(I) The function \(f(z)=\dfrac{x^2-y^2+2ixy}{x^2+y^2}\) (\(z=x+iy\)) has a limit as \(z\to0\).
(II) The function \(f(z)=\begin{cases}\operatorname{Re}(z)/|z|, & z\ne0\\ 1, & z=0\end{cases}\) is continuous at \(z=0\).
Choose the correct answer:

• A. Only (I) is true
• B. Only (II) is true
• C. Both (I) and (II) are true
• D. Neither (I) nor (II) is true

Hint:

Check the limit along different paths approaching the origin.

Answer 1

The Correct Option is D

Step 1: The numerator \(x^2-y^2+2ixy\) equals \(z^2\), and the denominator \(x^2+y^2\) equals \(|z|^2=z\bar z\). So for \(z\neq0\), \[f(z)=\frac{z^2}{z\bar z}=\frac{z}{\bar z}.\]

Step 2: Write \(z=re^{i\theta}\), so \(\bar z=re^{-i\theta}\) and \(f(z)=z/\bar z=e^{2i\theta}\). As \(z\to0\) along the ray of angle \(\theta\), the value stays \(e^{2i\theta}\), which depends on \(\theta\): along \(\theta=0\), \(f\to1\); along \(\theta=\pi/2\), \(f\to e^{i\pi}=-1\). Since different directions give different limiting values, \(\lim_{z\to0}f(z)\) does not exist. Statement (I) is false.

Step 3: For (II), \(f(z)=\operatorname{Re}(z)/|z|=x/\sqrt{x^2+y^2}\) for \(z\neq0\). Along the positive real axis (\(y=0,x\to0^+\)), \(f=1\); along the imaginary axis (\(x=0\)), \(f=0\). These path limits disagree, so \(\lim_{z\to0}f(z)\) does not exist, and hence \(f\) cannot be continuous at \(z=0\) no matter what value \(f(0)\) is assigned. Statement (II) is false.

Step 4: Neither (I) nor (II) is true. \[\boxed{\text{Neither (I) nor (II) is true}}\]