According to Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta;$ so, $e^{2\pi i} = 1.$ And $\ln(1) = 0.$
On the other hand, $\ln(e^{i\theta}) = i\theta,$ so $\ln(e^{2\pi i}) = 2\pi i$.
Is $\ln(e^{2i\pi})$ equal to $0$ or $2\pi i$?
Asked
Active
Viewed 319 times
7
-
Welcome to MSE! Good first question. – H Huang Oct 31 '21 at 07:55
-
1What do you mean by $\ln z$ for a complex number $z$? – Kavi Rama Murthy Oct 31 '21 at 07:55
-
2$2\pi i+2n\pi i$ is a logarithm of $e^{2\pi i}$ for every integer $n$. – Kavi Rama Murthy Oct 31 '21 at 07:56
-
3Congratulations! You've discovered that complex logarithm is multi-valued. – lisyarus Oct 31 '21 at 08:02
-
Very good question, you should study about Complex Logarithms. And you will be able to answer your question yourself. Happy learning! – prAnjal Oct 31 '21 at 08:19
-
Thanks! I just understood the $e^{i\theta}$ is $2 \pi$ periodic! – juzmathnerd Oct 31 '21 at 08:21
-
You can view $\ln$ as a (holomorphic) function on the universal covering space of $\mathbb{C}\setminus \left{ 0\right}$. Then $e^{2\pi i}$ is a unique point of that space for which $\ln(e^{2\pi i})=2\pi i$. See https://en.wikipedia.org/wiki/Complex_logarithm#The_associated_Riemann_surface – Gary Oct 31 '21 at 08:44
2 Answers
2
The natural logarithm is multivalued with respect to complex numbers. This means that the natural logarithm is not a function because its outputs are not unique. In order to deal with this, mathematicians tend to view what they call the "principal branch" of the natural logarithm or the main branch, denoted "$\operatorname{Ln} x$". While $\ln x$ is not a function, $\operatorname{Ln} x$ is. Therefore $\ln 1 = 0, 2\pi i,\dots$. In fact, it is equal to $2\pi in$ where $n$ is an integer all at the same time. It's not just 1 that has this property; all complex values for which the natural logarithm is defined share this property. However, given how it is generally defined, $\operatorname{Ln} x = 0$ only. If you know about $\arcsin$ and $\arccos$, it's kind of like that.
Great question. Let me know if I wasn't clear or if I got something wrong.
Also, as a side note, when using the $ for the math editing tool, you can write "\ln" in order to make it straight.
William Ryman
- 548
-
1Thanks for the answer! So the way is to do the logarithm is to see the $\ln x$ as a whole but not directly evaluate $x$ then $\ln x$ – juzmathnerd Oct 31 '21 at 08:20
-
No problem. It was a great question. I'm sorry, but I don't seem to understand what you are asking now. – William Ryman Oct 31 '21 at 08:24
-
1@juzmathnerd To compute $\ln(e^{2i\pi})$ (there will be infinitely many values), use definition (45) here. For example, the second equals sign in this Answer is an application of this definition to compute $\ln(i).$ – ryang Oct 31 '21 at 08:27
1
Please consider the fact that \begin{equation} \log z = \log |z|+j (arg(z)+2n \pi) \end{equation} where $z \in \mathbb{C}$, and $n=0,\pm 1,\pm 2, ...$, and $j=\sqrt{-1}$. As you wrote $|e^{2 \pi j}|=1$, while $arg(e^{2 \pi j})=0$. The complex argument can be computed by using the arc tangent function of the ratio between imaginary part and real part of the complex number. When doing so we can conclude that \begin{equation} \log e^{2 \pi j} = 2n \pi j \end{equation} When n = 0 we refer to as the principal branch of the logarithm; the corresponding value of the function is referred to as the principal value. If z is a real and positive number, we usually take $n = 0$ so that the principal branch of the complex logarithm function agrees with the usual one for real variables. In reality the function $\log z$ is infinitely branched.
Upax
- 2,073
- 13
- 18