8
$\begingroup$

I was playing with Wolfram|Alpha (just for fun, cause math is fun 😊), and I found that $\ln(-1)=i\pi$ which is quite obvious when you already know Euler's Identity. So, I continued playing with numbers and when I typed

$$\frac{2\ln(i)}{i}$$

it returned "$\pi$". And I have no idea why this is true. I looked it up on the internet, but I have not found anything interesting. Do you guys have a nice explanation for this?

$\endgroup$
  • $\begingroup$ Search for complex logarithm $\endgroup$ – Jannes Braet May 23 '17 at 7:57
  • $\begingroup$ $\ln z=\ln |z|+i\arg(z)$, where $\arg$ represent the principal argument of $z$, that is, the angle for $z$ in $(-\pi,\pi]$. Because $i$ is represented in the vertical axis of the plane of complex numbers then it principal argument is $\pi/2$. $\endgroup$ – Masacroso May 23 '17 at 8:07
  • 1
    $\begingroup$ $+1$ for writing "math is fun". It is also beauty ! $\endgroup$ – Claude Leibovici May 23 '17 at 8:15
2
$\begingroup$

Actually, the form $e^{i\pi }+1=0$ is just a special case of Euler 's identity, which states that $$e^{iz}=\cos{z}+i\sin{z}$$ So by Euler 's identity, $$e^{i\frac{\pi}{2}}=i$$ So that $$\frac{2\ln(i)}{i}=\frac{2i\frac{\pi}{2}}{i}=\pi$$

$\endgroup$
2
$\begingroup$

For the principal branch of the logarithm we have $\ln i=i\pi/2$, since $e^{i\pi/2}=i$. Hence $2\ln i=i\pi$, and upon dividing by $i$ the result follows.

$\endgroup$
2
$\begingroup$

This is equivalent to $$\ln(i) = i\frac{\pi}{2}$$ Taking $e$ to the power of both sides gives us $$e^{\ln(i)} = e^{i\frac \pi 2}$$ Both sides are equal to $i$ (the right side is because of Euler's formula, which states that for real $x$ we have $e^{ix} = \cos x+i\sin x$). So, everything works out.

$\endgroup$
2
$\begingroup$

Think of the logarithm as giving you back the angle between the positive real axis and the number you have plugged in (times a factor $i$). You were not surprised to see

$$\log(-1)=\pi i.$$

However, this statement just means that the angle between the positive axis, and the negative axis (represented by $-1$) is $180^°$ or (in radians) $\pi$. And when you look at the complex plain, you would see that $i$ is "above" the zero, hence there is a $90^°$ angle between the positive axis and the complex axis (represented by $i$), which expressed by radians is $\pi/2$. Hence

$$\log(i)=\frac\pi 2 i.$$

There is some more math and some more subtleties hidden inside. But this should suffice as a first explanation.


Maybe also this might be interesting. You probably know that logarithms "convert" multipication into addition in the sense $\log(ab)=\log(a)+\log(b)$. And you you know that $i\cdot i=-1$. So it would be quite natural to assume that

$$\log(i)+\log(i)=\log(-1).$$

And thats exactly what you found.

$\endgroup$
1
$\begingroup$

As $\operatorname{Ln}(-1)=i\pi$, $\operatorname{Ln}(i^2)=i\pi$, so $2\operatorname{Ln}(i)=i\pi$, and $\frac{2\operatorname{Ln}(i)}{i}=\pi$.

This uses the single-valued complex logarithm, $\operatorname{Ln}$. (see The complex logarithm, exponential and power functions (43), and equation (56), where we assume the principal value.)

$\endgroup$
  • 1
    $\begingroup$ in general $\ln z^n\neq n\ln z$ for $z\in\Bbb C$. In fact $n \ln i\neq\ln(i^n)$ for $n>2$. $\endgroup$ – Masacroso May 23 '17 at 10:22
  • $\begingroup$ @Masacroso; scipp.ucsc.edu/~haber/ph116A/clog_11.pdf has a decent discussion of ln vs. Ln, where Ln is single valued. I've edited my answer to reflect this $\endgroup$ – JMP May 23 '17 at 10:50
  • $\begingroup$ mmm... but how is this related to my previous comment? The identity $n \ln z=\ln z^n$ doesnt hold in general for the single valued complex logarithm or the set valued complex logarithm. In particular my comment is referred to the single valued complex logarithm aka principal value of the complex logarithm. $\endgroup$ – Masacroso May 23 '17 at 10:53
  • $\begingroup$ the reason why $2\ln i=\ln i^2$ is because the principal argument of $\ln i=\pi/2$ and $2\cdot\pi/2\in(-\pi,\pi]$. But you cant know that it holds if you dont know previously that $\ln i=\pi/2$ so the pass of the power out of the logarithm is not correct regardless that the result is right. $\endgroup$ – Masacroso May 23 '17 at 11:00
  • $\begingroup$ I mean $\arg(i)$ instead of $\ln i$ in my previous comment, sorry. $\endgroup$ – Masacroso May 23 '17 at 11:10
0
$\begingroup$

That is an example of multivalued functions in Complex Analysis. Assuming you know that every complex number $z$ can be written like $z=|z|e^{Argz}$, then $Argz$ is called $z$'s argument, which can be $\alpha, \alpha+2\pi,...,\alpha+2k\pi$. Then we define the first one argunent, namely $\alpha$, to be the main value of $Arg z$, named as $argz$. $Ln(z)$, just like $Argz$ is also a multivalued function in Complex Analysis. With simple calculation, we have $Ln(z)=ln|z|+iArgz$, where the function $ln(*)$ means the real logarithmic function. Now $Ln(i)=Ln(1*e^{\pi/2})=ln1+i(\pi/2+2k\pi)$, and we now define the main value of the multivalued function $Ln(*)$, named again as $ln(*)$, and $ln(z)=ln|z|+iarg(z)$. Generally speaking, we restrict $\alpha=argz\in (-\pi, \pi]$, and this ensures the uniqueness of the main value. So $ln(i)=0+i*(\pi/2)=i\pi/2$. So everything clear.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.