Question

If $f(z) = \frac{z}{\bar{z}}$, then $\lim_{z \to 0} f(z)$

• A. exists and equal to 0.
• B. does not exist.
• C. exists and equal to 1.
• D. exists and equal to $-1$.

Hint:

In complex numbers, $z/\bar{z} = e^{2i\theta}$ in polar coordinates $z = r e^{i\theta}$. As $r \to 0$, $f(z) = e^{2i\theta}$, which depends on the angle of approach $\theta$. Therefore, the limit at $z=0$ fails to exist!

Answer 1

The Correct Option is B

Step 1: Concept
In complex analysis, the limit $\lim_{z \to z_0} f(z) = L$ exists if and only if $f(z)$ approaches the exact same value $L$ along every possible path in the complex plane as $z \to z_0$. If approaching along two different straight line paths yields different values, the limit does not exist.

Step 2: Key Formulas and Approach
Let $z = x + iy$, so that its complex conjugate is $\bar{z} = x - iy$. The function becomes: \[ f(z) = \frac{x + iy}{x - iy} \] We test the limit as $(x, y) \to (0, 0)$ along straight line paths passing through the origin $y = mx$.

Step 3: Step-by-step Explanation

• Substitute $y = mx$ into $f(z)$: \[ f(z) = \frac{x + i(mx)}{x - i(mx)} = \frac{x(1 + im)}{x(1 - im)} = \frac{1 + im}{1 - im} \]
• Evaluate the limit along the path $y = mx$ as $x \to 0$: \[ \lim_{\substack{(x, y) \to (0, 0) y = mx}} f(z) = \frac{1 + im}{1 - im} \]
• Notice that this limiting value explicitly depends on the slope $m$ of the chosen path: - Along the real axis ($y = 0$, so $m = 0$): \[ \lim_{x \to 0} f(x) = \frac{1 + 0}{1 - 0} = 1 \] - Along the imaginary axis ($x = 0$): \[ \lim_{y \to 0} \frac{iy}{-iy} = -1 \] - Along the diagonal line $y = x$ ($m = 1$): \[ \lim_{x \to 0} \frac{1 + i}{1 - i} = \frac{(1+i)^2}{2} = i \]
• Since the limit depends on the path of approach ($1 \neq -1 \neq i$), the limit does not exist.

Step 4: Final Answer
Because different paths to the origin yield different limiting values, $\lim_{z \to 0} f(z)$ does not exist. Thus, Option (B) is correct.