Calculator test

[Commie]

(12-11-2024 08:12 PM)Idnarn Wrote:  The above is missing a term:

\[2k\pi i\]

See: https://en.wikipedia.org/wiki/Complex_logarithm
\[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\]

Now back to the original problem, The supposedly missing term turns out to be 'sine circles' were k denotes the wave number of the cycles, \[2k\pi i\] which should be written \[2\pi{k}{i}\]notice, it's a complete wave i.e., \[2\pi\]

At the end of the day, our calculators discard sine circles, which was the answer I was looking for.

Also, who do think writes the articles for Wikipedia?

Cheers
Darren

[Commie]

(12-12-2024 06:22 PM)AnnoyedOne Wrote:  2.pi is the cycle length for trigonometric functions but -pi/2 to pi/2 works better for some purposes. Its that simple.

Can you elaborate a bit more?

(12-12-2024 06:22 PM)AnnoyedOne Wrote:  Most simply assume that k=1 and move on.

eh?

Cheers
Darren

[AnnoyedOne]

(12-12-2024 09:14 PM)Commie Wrote:  Can you elaborate a bit more?

If you look at trigometric functions they repeat every 2.pi (or 360 deg). Going from -pi/2 to pi/2 gives their "negative" values although the 0 to pi/2 ones are the same (except for the sign). This is what I meant by cyclical and symmetric.

If you're an EE you can see this for yourself on an oscilloscope. No math required. Just eyes.

Assuming that k=1 (as calculators do) removes it from the equations.

That is 2.pi.k.i becomes 2.pi.i

A1

[Commie]

(12-13-2024 01:11 PM)AnnoyedOne Wrote:  Assuming that k=1 (as calculators do) removes it from the equations.

That is 2.pi.k.i becomes 2.pi.i

A1

But Naddy reckons k can be any integer, how do know which one to use?

naddy Wrote:k=…,−2,−1,0,1,2,…
; i.e., k∈Z
All of those (infinitely many) values are a correct solution for ln(reiθ)
. Many complex functions have more than one value for any argument. Typically one is designated the principal value, which in this case is the one you gave a formula for.

So, it turns out, I was not wrong?

[AnnoyedOne]

(12-13-2024 02:37 PM)Commie Wrote:  ...how do know which one to use?

Because k=1 is both valid and easy It also results in the "principal value" as the answer.

A1

PS: Think of 'k' as the cycle number going off to infinity in both direction. k=1 is end of the "first" cycle.

PPS: This is an image of a sinewave on an oscilloscope. See the pattern? Each up/down around zero is a "cycle".

https://i.ytimg.com/vi/2ect8INTebU/maxresdefault.jpg

[Albert Chan]

(12-13-2024 02:42 PM)AnnoyedOne Wrote:  Because k=1 is both valid and easy It also results in the "principal value" as the answer.

I think it is k=0, to get a nice transition to real ln, say ln(2) ≈ 0.693, without imaginary part.

The reason we like principal value is we have bijective function (one-to-one correspondence).

[AnnoyedOne]

(12-13-2024 06:10 PM)Albert Chan Wrote:  I think it is k=0, to get a nice transition to real ln, say ln(2) ≈ 0.693, without imaginary part.

I think that that this thread is about ln(-1) the result of which is complex but has an "infinite" number of answers (with 'k').

A1

[Commie]

(12-13-2024 06:10 PM)Albert Chan Wrote:  

(12-13-2024 02:42 PM)AnnoyedOne Wrote:  Because k=1 is both valid and easy It also results in the "principal value" as the answer.

I think it is k=0,

You are correct Albert, otherwise the imaginary part 'jumps' or starts from to 2.pi.i

A1, what about values below 2.pi?

ah,ha we getting somewhere, at last.

Cheers
Darren

[AnnoyedOne]

(12-13-2024 06:30 PM)Commie Wrote:  ...what about values below 2.pi?

I'm not sure what you mean. Are you referring to the 2.pi.i component? If so that is the answer.

A1

[Commie]

(12-13-2024 06:22 PM)AnnoyedOne Wrote:  I think that that this thread is about ln(-1) the result of which is complex but has an "infinite" number of answers (with 'k').

A1

No, we are talking about the complex logarithm, please keep on topic.

Darren

[Commie]

(12-13-2024 06:34 PM)AnnoyedOne Wrote:  I'm not sure what you mean. Are you referring to the 2.pi.i component? If so that is the answer.

A1

How does ln(-1)=pi.i?, if you start with i=2.pi.i with k=1

How does the imaginary i part of the complex logarithm result equate to values below 2.pi if k=1?


Cheers
Darren

[Commie]

Hi A1 and Naddy,

In the following equation : \[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\] stated by Idnarn, says that, if k=1 as stated as the principal value by Naddy and A1, then if theta is zero, then i=2.pi. The question then arises, how does the imaginary part of Ln(z) take on values less than 2.pi.?

If theta = pi, then according to Naddy and A1, the imaginary part = i.(pi+2.pi) with k=1 right?

I look forward to your reply and btw A1, I do know what a sinewave looks like on an oscilloscope.

[naddy]

(12-13-2024 09:46 PM)Commie Wrote:  Hi A1 and Naddy,

In the following equation : \[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\] stated by Idnarn, says that, if k=1 as stated as the principal value by Naddy and A1,

I said no such thing. The principal value is \[\ln(r e^{i \theta}) = \ln r + \ln(e^{i \theta}) = \ln r + i \theta\] I think definitions are generally chosen such that you have the principal value at \(k = 0\).

[Commie]

(12-13-2024 10:45 PM)naddy Wrote:  I said no such thing. The principal value is \[\ln(r e^{i \theta}) = \ln r + \ln(e^{i \theta}) = \ln r + i \theta\] I think definitions are generally chosen such that you have the principal value at \(k = 0\).

Hi Naddy,

This what you posted,

(12-11-2024 11:14 PM)naddy Wrote:  \(k = \dots, -2, -1, 0, 1, 2, \dots\) ; i.e., \(k \in \mathbb Z\)

All of those (infinitely many) values are a correct solution for \(\ln(r e^{i\theta})\). Many complex functions have more than one value for any argument. Typically one is designated the principal value, which in this case is the one you gave a formula for. Section 3 of the HP-15C Advanced Functions Handbook has some helpful notes about this.

Think square roots: \(\sqrt 4 = \pm 2\), but the calculator will only give you the principal value, \(2\).

And, here is what A1 posted...

(12-13-2024 02:42 PM)AnnoyedOne Wrote:  Because k=1 is both valid and easy It also results in the "principal value" as the answer.

Cheers
Darren

[EdS2]

I think it's better to seek clarification, than to exchange blunt words.

It's also better to wait a bit before responding. Wiser heads might have better responses but not be here all the time.

Otherwise we get an escalation, and we get opinions or guesses treated as facts.

Slow down, calm down. This is not about scoring points.

[AnnoyedOne]

(12-13-2024 10:45 PM)naddy Wrote:  I think definitions are generally chosen such that you have the principal value at \(k = 0\).

You (and Albert Chan) are probably correct about that. I forgot about the theta.i term which makes any result complex (vs real) except if theta=0 in which case you get ln(r). k=0 makes things even easier

This stuff is often drawn in 3D so that the 'i' value is simply a rotation about the y-axis (i.e. is on the z-axis). 2.pi is 360 degrees (a full rotation) which gets you back to where you started (just as with trig functions). That is why the value of 'k' really isn't important. Any integer will work. Negative ones just mean rotating in the other direction.

A1

[naddy]

(12-14-2024 05:36 AM)Commie Wrote:  This what you posted,

(12-11-2024 11:14 PM)naddy Wrote:  All of those (infinitely many) values are a correct solution for \(\ln(r e^{i\theta})\). Many complex functions have more than one value for any argument. Typically one is designated the principal value,

Yes and you seem to have misunderstood me. Let's clarify:

Quote:Many complex functions have more than one value for any argument. Typically one [of those values] is designated the principal value,

[AnnoyedOne]

(12-14-2024 02:15 PM)naddy Wrote:  Typically one [of those values] is designated the principal value,

I never bothered looking up the definition but here it is

https://en.wikipedia.org/wiki/Principal_value

Quote:The branch corresponding to k = 0 is known as the principal branch, and along this branch, the values the function takes are known as the principal values.

A1

[Commie]

Hi Chaps,

The correct equation for Ln(z)=Ln(x+iy) is:
\[ln({x+iy}) = \frac{1}{2}ln({x^{2}+y^{2}})+{i}arctan(\frac{y}{x}) \pm2k\pi i\]

Consider, you have z=3-4i. we have Ln(3-4i)=1.6094-0.9273i+0i , k=0

We can take e^(1.6094-0.9273i) = 3-4i back where we started.

If we now increment/decrement it's imaginary part by 2.pi then

Ln(3-4i)=1.6094-0.9273i+2.pi.i= 1.6094+5.3559i , k=1
we can take e^(1.6094+5.3559i)= 3-4i back where we started.

Now increment/decrement it's imaginary part by another 2.pi then

Ln(3-4i)=1.6094-0.9273i+4.pi.i= 1.6094+11.6391i k=2
We can take e^(1.6094+11.6391i)= 3-4i back where we started.

When k=0, the imaginary is in ground state, i.e., the smallest possible(principal) value for the imaginary i.

And so on, these imaginary 2.pi jumps exist between +- infinity. However, no matter how large the imaginary grows in 2.pi increments or decrements, it will always yield, in the example shown, (3-4i) back where we started.

The real always stays constant, whilst the imaginary i is quantized in steps of 2.pi.

Stay safe guys
Darren

[Thomas Klemm]

Just nitpicking:

• use \ln and \arctan for functions
• use \left( and \right) for nested parenthesis
  

\[\ln({x+iy}) = \frac{1}{2}\ln\left({x^{2}+y^{2}}\right)+{i}\arctan\left(\frac{y}{x}\right) \pm2k\pi i\]