How Can \dot{\gamma}(0) Fail to Exist in Palka's Example 1.3?

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SUMMARY

The discussion centers on the failure of the derivative $$\dot{\gamma}(0)$$ to exist in Example 1.3 of Bruce P. Palka's "An Introduction to Complex Function Theory." The analysis reveals that the difference quotient $$\dfrac{\gamma(t) - \gamma(0)}{t}$$ yields different values for $$t < 0$$ and $$t > 0$$, specifically $$1-i$$ and $$1+i$$, respectively. Consequently, since the limit as $$t \to 0$$ cannot converge to a single value, it is established that $$\dot{\gamma}(0)$$ does not exist, confirming that the path $$\gamma$$ is not smooth.

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I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter 4: Complex Integration, Section 1.2 Smooth and Piecewise Smooth Paths ...

I need help with some aspects of Example 1.3, Section 1.2, Chapter 4 ...

Example 1.3, Section 1.2, Chapter 4 reads as follows:View attachment 7419In the above text from Palka, we read the following:

"... ... Since $$\dot{ \gamma } ( 0 )$$ fails to exist, $$\gamma$$ is not smooth ... ... "I wish to rigorously demonstrate that $$\dot{ \gamma } ( 0 )$$ fails to exist ... but do not know how to proceed ...

Can someone please show me how to rigorously prove that $$\dot{ \gamma } ( 0 )$$ fails to existHelp will be much appreciated ...

Peter
 
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Form the difference quotient $\dfrac{\gamma(t) - \gamma(0)}t$. If $t<0$, this is $1-i$. If $t>0$, it is $1+i$. The derivative $\dot{\gamma}(0)$, if it exists, is the limit as $t\to0$ of the difference quotient. If the limit exists, it would have to be both $1-i$ and $1+i$. So the conclusion is that it does not exist.
 
Last edited:
Opalg said:
Peter said:
Form the difference quotient $\dfrac{\gamma(t) - \gamma(0)}t$. If $t<0$, this is $1-i$. If $t>0$, this is $1+i$. The derivative $\dot{\gamma}(0)$, if it exists, is the limit as $t\to0$ of the difference quotient. If the limit exists, it would have to be both $1-i$ and $1+i$. So the conclusion is that it does not exist.


Thanks Opalg ...

... straightforward when you see how ...

Peter
 

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