Question:medium

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

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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!

Updated On: Jul 29, 2026

exists and equal to 0.
does not exist.
exists and equal to 1.
exists and equal to $-1$.

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The Correct Option is B

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Top Questions on Complex Analysis

Match List-I with List-II and choose the correct option:

LIST-I (Function)	LIST-II (Value)
(A) $ \int_{\gamma} \frac{1}{z-a} \, dz $, where $ \gamma: \|z-a\|=r, r > 0 $	(III) $ 2i\pi $
(B) $ \int_{\gamma} \frac{z+2}{z} \, dz $, where $ \gamma: z = 2e^{it}, 0 \le t \le \pi $	(IV) $ i\pi $
(C) $ \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz $, where $ \gamma: \|z\|=3 $	(II) $ 2i\pi(e^4 - e^2) $
(D) $ \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz $, where $ \gamma: \|z\|=2 $	(I) $ -4 + 2i\pi $

Choose the correct answer from the options given below:

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LIST-I (Function)	LIST-II (Value)
(A) \( \int_{\gamma} \frac{1}{z-a} \, dz \), where \( \gamma: \|z-a\|=r, r > 0 \)	(III) \( 2i\pi \)
(B) \( \int_{\gamma} \frac{z+2}{z} \, dz \), where \( \gamma: z = 2e^{it}, 0 \le t \le \pi \)	(IV) \( i\pi \)
(C) \( \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz \), where \( \gamma: \|z\|=3 \)	(II) \( 2i\pi(e^4 - e^2) \)
(D) \( \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz \), where \( \gamma: \|z\|=2 \)	(I) \( -4 + 2i\pi \)