Complex Log In!

Algebra Level 2

Evaluate the complex logarithm:

$\large \log_{-6} -6 =$

Not defined 2 0 1

4 solutions

Aditya Narayan Sharma
Mar 30, 2016

$log_{-6}-6=\frac{log_{10}(-6)}{log_{10}(-6)}$

Let $log_{10}(-6)=x\implies 10^x=-6$ , But any power of 10 can't be negative. Hence it's undefined

Complex logarithms exist, but are interpreted as multi-valued functions.

It's tough to judge the context of this problem, and for people who are studying logarithms over the (positive) reals, this doesn't exist. As such, I have clarified that the context, that we should treat this as a complex logarithm.

Calvin Lin Staff - 5 years, 2 months ago

then, Calvin wich is the complex logarithm $\ln e$ ? $1$ or $1 + 2i \pi$ or $1 + 4i \pi$ ... or? and in this case, which is $\log_{-6} -6$ ? because $-6 = (-6)^1 =(-6)^{e^{2i \pi}}=(-6)^{ e^{4i\pi}}$ ? don't you believe,in this case, if we don't stablish the domain and range of a continuous "log" the answer is not defined?I'm only asking, I'm very curious...

Guillermo Templado - 5 years, 2 months ago

As I stated, they are multi-valued functions, and so they have their own interpretations which could differ from your intuitive understanding of functions.

If $f$ is a multi-valued function, then we say that $f(a) = b$ if $b$ is one of the possible values of $f(a)$ .

The idea of continuity can also be extended, namely, a multi-valued function is continuous at $f(a) = b$ , if there exists a (single valued) function $f_b (x)$ in the neighbourhood of $(a,b)$ such that $f(x) = f_b (x)$ and $f_b(x)$ is a continuous function.

As an example, the argument function on the complex plane, can be defined as a multi-valued function where the values differ by multiples of $2 \pi$ . Then, we can see that $\arg$ is continuous at all points other than $\arg (0)$ (which is a reason why it's often left as undefined).

Calvin Lin Staff - 5 years, 2 months ago

@Calvin Lin – Then,Are you saying me then for example, that multivalued function $x^{\frac{1}{3}}$ on and over the complex numbers and what if I take one single valued function of these functions is continuous in a neighborhood of 0 ?

Guillermo Templado - 5 years, 2 months ago

@Guillermo Templado – The multivalued function $f: \mathbb{C} \rightarrow \mathbb{C}$ , $f(x) = x ^ { \frac{1}{3} }$ is continuous at every point.

Typically, we like to stay away from "branch" points, because it leads to messiness. It is nicer to consider the function as a "stack of pancakes" (manifold), where the restriction to each layer keeps things nice.

For example, the inverse function theorem also extends to manifolds as we would expect it to.

Calvin Lin Staff - 5 years, 2 months ago

@Calvin Lin – Sorry, but the maltivalued function $f: \mathbb{C} \rightarrow \mathbb{C}$ $f(x) = x^\frac{1}{3}$ is not continuous in any neighborhood of 0, and I'm going to say why? Let the circumference $\{x \in \mathbb{C} \space / |x| = r\} \space r \in \mathbb{R}, \space r > 0$ is contained in a neighborhood of 0. Let $x = r \cdot e^{i\theta}$ then if we walked a loop around this circumference (spin along) the circunference we would get $r^{\frac{1}{3}} = r^{\frac{1}{3}} \cdot e^{\frac{2i \pi}{3}}$ which it is not true, for that this function can't be continuous in any neighborhood of 0, and for that, althought you can take the principal logarithm to be the most beautiful and clear logarithm for this problem, you can cause ambiguety to the people, like in this case to me, althought I don't mind my level and credits, I only mind everything is allright.

Guillermo Templado - 5 years, 2 months ago

@Guillermo Templado –

I do not see what continuity of a multivalued function (complex logarithm) has to do with this problem. Note that you are in the solution discussion, which should be kept relevant to the solution at hand.
In the space of a comment, I am not writing out a wiki page with all of the necessary knowledge to understand multi-valued functions. I am assuming (as with everything else on the site) that if you are interested in knowing this further, you will do some work on your own.
Continuity at a point does not imply that the (restricted) function is continuous on any other point in the neighbourhood. At $x = 0$ , I can still define $f_0 (x)$ to make the restricted function continuous at $0$ , even though it will not be continuous at any non-zero real. You can prove this because if we take any sequence $x_n$ that tends to zero, it is clear that $x_n ^ { \frac{1}{3} }$ also tends to 0.
Avoid conflating "function is continuous at a point" with "function is continuous everywhere". Especially for multivalued functions like you see.

If you are interested in understanding further, pick up a book analysis on manifolds.

Calvin Lin Staff - 5 years, 2 months ago

@Calvin Lin – Then if $\{x \in \mathbb{C} \space / |x| = r \space r \in \mathbb{R}, \space r > 0\} \subset V$ $f(x) = x^{\frac{1}{3}}$ is continuous in $V$ . Great .... don't worry, I have and picked up a lot of books about manifolds and generalized surfaces.... and my question is why $log_{-6} -6 = 1$ and it isn't infinite values, I mean $-6 = 6 \cdot e^{i\pi} = 6 \cdot e^{3i \pi}$ , then what is $\ln (-6) = \ln |6| + i\pi$ or $\ln (-6) = \ln |6| + 3i \pi$ ?and then what is $\log_{-6} e$ ?... and then what is $\log_{-6} -6$ ?

Guillermo Templado - 5 years, 2 months ago

@Guillermo Templado – As I stated above:

As I stated, they are multi-valued functions, and so they have their own interpretations which could differ from your intuitive understanding of functions.

If $f$ is a multi-valued function, then we say that $f(a) = b$ if $b$ is one of the possible values of $f(a)$ .

So, we say that $\log_{-6} -6 = 1$ because $(-6) ^ 1 = -6$ . Similarly, we can say that $\ln 1 = 0$ and $\ln 1 = 2 \pi i$ and $\ln 1 = - 2 \pi i$ . However, this does not imply that $0 = 2 \pi i$ .

Simlarly, when dealing with the multi-valued "square root", even though we want to say that $\sqrt{1} = 1$ and $\sqrt{1} = -1$ , we cannot conclude that $1 = -1$ (which is the common way to arrive at the fallacy.

Calvin Lin Staff - 5 years, 2 months ago

But any power of 10 can't be negative.

They can be.

Precisely, $\log_{10} (-6) = \frac{i \pi + \ln 6}{\ln{10}}$

Agnishom Chattopadhyay - 5 years, 2 months ago

This is true. But if we are dealing it on the real plane, it won't exist.

Aditya Kumar - 5 years, 2 months ago

The question does not state that. and more than that, I do not see any reason to be restricting to reals!

Also, rearranging a form to a form that makes it undefined is not a valid step.

For example, is the following valid?

$0^1 = 0^{2-1} = \frac{0^2}{0}$

Agnishom Chattopadhyay - 5 years, 2 months ago

@Agnishom Chattopadhyay – Hmm yes you are right. Thanks for clearing.

Aditya Kumar - 5 years, 2 months ago

Yeah ofcrs i agree with you. But as the classical definition of log states that negative base is undefined while dealing for reals. I have just extended that idea in he real field

Aditya Narayan Sharma - 5 years, 2 months ago

Aniruddha Bagchi
Apr 2, 2016

Hey, it's simply 1.The answer is 1.(-6)^1=(-6).

Mathieu Roy
Apr 8, 2016

It is simply asking what power can you give to (-6) to get (-6). Naturally the answer will be 1.

A Former Brilliant Member
Apr 3, 2016

Simple way:

$\log_{-6}(-6) = 1 \qquad{\text{Apply the rule: } \log_{a}(a) =1 }$

Complex Log In!

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