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I just watched a video lecture where it's proved that $\lim_{z\rightarrow 0}{\frac{\bar{z}}{z}}$ does not exist: by putting the condition that $z=x+iy$ and $y=\lambda x$, it comes out that $\frac{\bar{z}}{z}=\frac{1-i\lambda}{1+i\lambda}$, which depends on $\lambda$, hence the limit doesn't exist.

However WolframAlpha says it's equal to 1. Why? Is it making some more advanced assumptions? Is the proof above correct?

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Asking it to show steps, it seems to think the derivative of conjugation converges to 1 as z goes to 0. –  Hurkyl Mar 31 '12 at 10:19

2 Answers 2

up vote 1 down vote accepted

The limit $\lim_{z\to 0}{\frac {\bar z} z}$ doesn't exist. Maybe some more specific examples will be more convincing a proof ($t\in \mathbb R$):

$$z=t+it \implies z\to 0 \iff t\to0 \implies \lim_{z\to 0}{\frac{\bar z}z}= \lim_{t\to0}{\frac{t-it}{t+it}}= \frac{1-i}{1+i} = -i$$ $$z=it \implies z\to 0 \iff t\to0 \implies \lim_{z\to 0}{\frac{\bar z}z}= \lim_{t\to0}{\frac{-it}{it}}= -1$$ $$z=t \implies z\to 0 \iff t\to0 \implies \lim_{z\to 0}{\frac {\bar z} z}= \lim_{t\to0}{\frac{\bar t}{t}}= \lim_{t\to0}\frac{t}{t} = 1$$

Not sure what WolframAlpha is doing, presumably it's not considering $z$ to be a complex variable, which would coincide with the third example above.

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Wolfram gives $\{-1,1\}$, when you ask him to calculate $$ \lim_{a\to0, b\to 0} \frac{a+ib}{a-ib} $$

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