Why does ln(i) = (1/2pi)i?

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Why does ln(i) = (1/2pi)i????

I was bored the other day and wondered whether or not it would be possible to find out the natural log of the imaginary number i. Typed it into my TI-84 and it said the answer was 1.57079632i. I wondered why the might be the case, thought about it for a while and noticed that 1.5707 is equal to 1/2pi. Decided to ask my calculator what ln(-1) equaled, which was as predicted pi*i.

ln(i) = 1/2ln(-1)
ln(-11/2) = 1/2ln(-1)
1/2ln(-1) = 1/2ln(-1)

Makes enough sense to me.

What confuses/interests me is this: What is it that ties the number i, the natural logarithm, and pi together? How does 1/2pi(i) get spit out? I had no idea that pi and i could possibly be related to each other in any way. Why is pi = [ln(1)/i]?

Further, can the perimeter/are of a circle be defined in terms of i?
 
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  • #2
arildno
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Hi, zktrouble!

The answer is that when we deal with complex&imaginary numbers, the exponential function and the trigonometric functions are closely related!

In fact, the complex exponential follows the following identity:
[tex]e^{i*x}=\cos(x)+i*\sin(x)[/tex] where x is measured in radians.
Thus, note that we have:
[tex]e^{i*\frac{\pi}{2}}=\cos(\frac{\pi}{2})+i*\sin(\frac{\pi}{2})=i[/tex], where it "follows" that the natural logarithm to i is [tex]i*\frac{\pi}{2}[/tex]
 
  • #3
Pengwuino
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Pick up a text on complex analysis. The logarithm function is typically only defined for positive real numbers. In order to extend the function to the entire complex plane except for the negative real numbers, you have to redefine the logarithm in such a way that it is a single-valued operation.
 
  • #4
arildno
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Pick up a text on complex analysis. The logarithm function is typically only defined for positive real numbers. In order to extend the function to the entire complex plane except for the negative real numbers, you have to redefine the logarithm in such a way that it is a single-valued operation.
I'm not sure who you are addressing. I know that, but didn't think it worth mentioning at the beginning, judging the level OP seemed to be in his knowledge of complex numbers.
 
  • #5
Pengwuino
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I'm not sure who you are addressing. I know that, but didn't think it worth mentioning at the beginning, judging the level OP seemed to be in his knowledge of complex numbers.
It was meant at the OP, I should have done a quote. I didn't get the same impression of the OP's familiarity with complex numbers and how the logarithmic function is not exactly what it is in the reals.
 
  • #6
arildno
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Well, OP knows best.
I was unsure if he knew of the Argand plane, and if he isn't, he'd need an introduction to that (and polar representation of a number) before understanding the geometrically very simple reason why complex logarithms are multi-valued.
 
  • #7
Pengwuino
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Well, OP knows best.
I was unsure if he knew of the Argand plane, and if he isn't, he'd need an introduction to that (and polar representation of a number) before understanding the geometrically very simple reason why complex logarithms are multi-valued.
I think you don't have to go that far. All you need is your example of why log(i) is what it is and you can extend that example to show they are multi-valued.

What I have in mind is the following:

[tex]e^{i {{\pi} \over {2}}} = e^{i {{\pi} \over {2}} + 2\pi i} = i[/tex]

Where it's fairly trivial to see that those two values are the same and the logarithm must be multi-valued if thought of in that way.
 
  • #8
arildno
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I think you don't have to go that far. All you need is your example of why log(i) is what it is and you can extend that example to show they are multi-valued.

What I have in mind is the following:

[tex]e^{i {{\pi} \over {2}}} = e^{i {{\pi} \over {2}} + 2\pi i} = i[/tex]

Where it's fairly trivial to see that those two values are the same and the logarithm must be multi-valued if thought of in that way.
It will depend upon his reaction.
If he wishes, for example, to understand why/I] there is such a close relationship between exponentiation and trigonometrics in the complex world, then understanding the definition behind complex multiplication and its relation to addition of angles will require an intro to the Argand plane.
 
  • #9
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arildno's first post makes enough sense, but I don't understand much about the other things everybody has been mentioning. My knowledge of complex numbers goes about as far as a college algebra course. Now I'm starting calculus but I haven't done much with imaginary/complex numbers in calculus yet. So what is this Argand plane because I haven't heard of it. Is this the same as plotting points on the graphs of the form (x, yi)?? Cheers.
 
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  • #10
arildno
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arildno's first post makes enough sense, but I don't understand much about the other things everybody has been mentioning. My knowledge of complex numbers goes about as far as a college algebra course. Now I'm starting calculus but I haven't done much with imaginary/complex numbers in calculus yet. So what is this Argand plane because I haven't heard of it. Is this the same as plotting points on the graphs of the form (x, yi)?? Cheers.
Yes, the Argand plane is the complex number plane.

In order to understand why we have the connection between the exponential and the trigonometrics, we can look at the following points:

1. An exponential function Exp(x) has the properties Exp(x)*Exp(y)=Exp(x+y) (and the correlate Exp(0)=1, if Exp is non-constant).
That is, multiplication of exponentials can be replaced by addition of exponents.

2. We have the trigonometric addition formula:
[tex]\cos(x)\cos(y)-\sin(x)\sin(y)=\cos(x+y), \sin(x)\cos(y)+\cos(x)\sin(y)=\sin(x+y)[/tex]

3. Multiplying two complex numbers, we have the general rule:
[tex](a+b*i)*(c+d*i)=(ac-bd)+(ad+bc)*i[/tex]

4. Now, pick two numbers, X and Y, both at unit distance from the origin in the complex number plane, and multiply them together. Their real and imaginary parts are then representable in terms of the cosine and sine functions. Let x and y be the angles associated with their positions in the complex plane, measured relative to the real axis.
[tex]X*Y=(\cos(x)+i*\sin(x))*(\cos(y)+i*\sin(y))=(\cos(x)\cos(y)-\sin(x)\sin(y))+(\sin(x)\cos(y)+\cos(x)\sin(y))*i=\cos(x+y)+\sin(x+y)*i[/itex]

That is, multiplying to complex number of unit length together is equivalent of adding their angles together!
Thus, angles in the complex plane fulfills the same function as the exponent in the exponential function.
 
  • #11
disregardthat
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If you have touched upon Taylor series in calculus, you will see that this extension of e^x is completely natural in the sense that the values of its Taylor series for complex input will coincide with the identity arildno mentioned. It would be a good exercise to verify this by comparing the Taylor series for e^x with the Taylor series for the sine and cosine function (without bothering with potential issues with convergence). That will hopefully clear up any confusing regarding the seemingly arbitrary definition of e^x for imaginary values.
 
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  • #12
arildno
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Now, to understand the idea of the complex logarithm as multi-valued:

Let z be a complex number of unit length, therefore we may write:
[tex]z=e^{i*\theta}=\cos(\theta)+i*\sin\theta, 0\leq\theta<{2\pi}[/tex]
Thus, trivially, we ought to have:
[tex]\log(z)=i*\theta[/tex]
Agreed?

But, and here's the thing:
Sines and cosines ae periodic functions; and we see that for every integer k, we could equally well say:
[tex]\log(z)=i*(\theta+k*2\pi)[/tex]
Thus, all complex numbers have infinitely many logarithms, and whenever we wish to refer to any single one of them, we have to specify which one.

The angular value that lies between 0 and two pi is generally called the principal value.
 
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  • #13
Delta2
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What confuses/interests me is this: What is it that ties the number i, the natural logarithm, and pi together? How does 1/2pi(i) get spit out? I had no idea that pi and i could possibly be related to each other in any way. Why is pi = [ln(1)/i]?
The answer is e, the base of the natural logarithms. There is the basic relation that
[tex]e^{i\pi}=-1=i^2[/tex]

which bounds e, pi, i and 1 ...This relation is quite "famous" , some even consider it as strong indication or even proof of god's existence . Of course objectively speaking it has nothing to do with god, it is just a nice relation...Proof of that relation and of the more general [tex]e^{ix}=cos(x)+isin(x)[/tex] follows from the taylor expansion of e^x, sinx and cosx.
 
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  • #14
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The answer is e, the base of the natural logarithms. There is the basic relation that
[tex]e^{i\pi}=-1=i^2[/tex]

which bounds e, pi, i and 1 ...This relation is quite "famous" , some even consider it as strong indication or even proof of god's existence . Of course objectively speaking it has nothing to do with god, it is just a nice relation...Proof of that relation and of the more general [tex]e^{ix}=cos(x)+isin(x)[/tex] follows from the taylor expansion of e^x, sinx and cosx.
e^(pi*i) = -1
ln (e^(pi*i)) = ln(-1)
pi*i = ln(-1)

i = ln(-1)/pi
AND
pi = ln(-1)/i

This ties things together quite nicely. Thanks!
 
  • #15
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Another idea:

Why does Summation from 0-->infinity of 1/n! = pi?
 
  • #16
arildno
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Another idea:

Why does Summation from 0-->infinity of 1/n! = pi?
It equals e, not pi.
 
  • #17
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It equals e, not pi.
Right, that's what I meant. not pi, e. But in any case, why does it?

[EDIT]

What is the magic involved in the summation of 1/n! that magically creates the base of the function which equals its own derivative?

[/EDIT]
 
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  • #18
arildno
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Right, that's what I meant. not pi, e. But in any case, why does it?
It really depends on your starting point.
Some might say that the number "e" is defined as that sum. In that case, there is no "why".
 
  • #19
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e is defined by that sum? I have always known e to be defined as follows:

Lim x--> infinity (1+1/n)^n. But I guess a summation to infinity is in essence a limit, so its understandable that perhaps the mentioned function can be simplified to the Sigma(1/n!)function. But how does it simplify?

Thanks again!
 
  • #20
arildno
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e is defined by that sum? I have always known e to be defined as follows:

Lim x--> infinity (1+1/n)^n. But I guess a summation to infinity is in essence a limit, so its understandable that perhaps the mentioned function can be simplified to the Sigma(1/n!)function. But how does it simplify?

Thanks again!
Those definitions can be proven to be equivalent.
It is an ugly and tedious business to prove that equivalence.

EDIT:

The beginning of the proof is quite simple:
By the binomial formula, you have, for integer "n":
[tex](1+\frac{1}{n})^{n}=\sum_{i=0}^{n}\binom{n}{i}\frac{1}{n^{i}}=\sum_{i=0}^{n}\frac{1}{i!}\Pi_{j=0}^{i-1}\frac{n-j}{n}[/tex]
Now, termwise, the pi products "ought" to converge to 1, and thus, the sum "ought" to converge to the infinite sum we want.
The drudgery is just to prove that.
 
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  • #21
HallsofIvy
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[tex]e= \lim_{n\to \infty} \left(1+ \frac{1}{n}\right)^n[/tex]
means that, for large n, [itex]\left(1+ \frac{1}{n}\right)^n[/tex] is very close to e.

From that,
[tex]1+ \frac{1}{n}\approx e^{1/n}[/tex]
[tex]e^{1/n}- 1\approx \frac{1}{n}[/tex]
Now let h= 1/n so we have
[tex]e^h- 1\approx h[/tex]

Now suppose we define [itex]f(x)= e^x[/itex]. To find the derivative we look at
[tex]\lim_{h\to 0}\frac{e^{x+h}- e^x}{h}[/tex]
[tex]= \lim_{h\to 0}e^x\frac{e^h- 1}{h}[/tex]
which because
[tex]\frac{e^h-1}{h}\approx\frac{h}{h}= 1[/tex]
becomes [itex]e^x(1)= e^x[/itex].

That is, with e defined as
[tex]\lim_{n\to\infty}\left(1+ \frac{1}{n}\right)^n[/itex]
we have
[tex]\frac{de^x}{dx}= e^x[/tex]

From that, it follows that [itex]e^x[/itex] is infinitely differentiable, that all derivatives are [itex]e^x[/itex] which has value 1 at x= 0, so that the MacLaurin series for [itex]e^x[/itex] gives
[tex]e^x= \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]
and, taking x= 1 in that,
[tex]e= \sum_{n=0}^\infty \frac{1}{n!}[/tex].
 
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