Author Topic: Section 1.4 Question 12  (Read 1038 times)

Heng Kan

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Section 1.4 Question 12
« on: October 02, 2018, 11:11:36 PM »
For this question, we have to compute the limit of $f(z) = (z-2)\log|z-2|$ at z=2. As z approaches to $2$, $z-2$ approaches to $0$ and $\log|z-2|$ approaches to negative infinity. If this is a real function, we can use the L'Hopital's rule to compute the limit. So in this situation where it is a complex function, how do we find the limit?  Thanks.
« Last Edit: October 03, 2018, 03:07:12 AM by Victor Ivrii »

Victor Ivrii

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Re: Section 1.4 Question 12
« Reply #1 on: October 03, 2018, 03:18:42 AM »
This is a tricky question, and most likely an erroneous, because $\log |z-2|$ is a multivalued function and we have not defined the limits of such functions. And depending on how you make a definition, you'll get different answers. Observe that this is the only problem with $\log$.

Therefore I slightly change the function to $f(z)= (z-2)\ln |z-2|$ (here $|z-2|$ is a real positive and $\ln $ is an ordinary logarithm). Try to write $(z-2)$ in the polar form.

oighea

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Re: Section 1.4 Question 12
« Reply #2 on: October 04, 2018, 01:47:33 AM »
Here is a major hint: log in some contexts may actually mean ln when the variables is real. Also, note that the absolute value |.| maps from $\mathbb{C}\mapsto \mathbb{R}^+$. Hence, log|z-2| is just a real logarithm, and is in fact defined everywhere on the complex plane except for z=2. This can be done without using the delta-epsilon definition of limits (which is unfortunately hard to verify).

We have the expression $\lim\limits_{z \to 2} (z-2) \ln|z-2|$. The lim is of the form $0\cdot-\infty$, meaning L'Hôpital's Rule applies.

Note that for L'Hôpital's Rule, $0 = \frac{1}{\infty}$. Hence, $(z-2) = \frac{1}{(z-2)^{-1}}$, and the lim of both when they approach 2 are zero.

$\lim\limits_{z \to 2} (z-2) \ln|z-2| = \lim\limits_{z \to 2}\frac{\ln|z-2|}{(z-2)^{-1}}$, which is of the $\frac{\infty}{\infty}$ form, the desired indeterminate form.

We apply L'Hôpital's Rule: $\lim\limits_{z \to 2}\frac{\ln|z-2|}{(z-2)^{-1}} \Rightarrow \lim\limits_{z \to 2}\frac{(z-2)^{-1}}{-1(z-2)^{-2}}=\lim\limits_{z \to 2}-\frac{(z-2)^{-1}}{(z-2)^{-2}}$. That looks very good, except the exponents are both negative, meaning that they switch places on the fraction bar.

We now arrive at $\lim\limits_{z \to 2}-\frac{(z-2)^{2}}{(z-2)^{1}}$, and finally $\lim\limits_{z \to 2}(z-2)$. We evaluate the limit directly: $(2-2) = 0$

∴ the lim of $(z-2) \ln|z-2|$ is zero.
« Last Edit: October 04, 2018, 01:53:18 AM by oighea »

Victor Ivrii

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Re: Section 1.4 Question 12
« Reply #3 on: October 05, 2018, 07:11:57 AM »
It is fine, however we do not have L'Hôpital's Rule for complex-valued functions. However, if we want to prove that something goes to $0$, we actually need to prove that it's modules goes to $0$. So in fact we need to prove that $|z-2|\ln |z-2|$ goes to $0$ as $z\to 2$, which means that we trying to prove that $r\ln r\to 0$ as $r\to 0$.