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AlbertEinstein Jun22-06 12:33 PM

To find logarithm of a complex number
 
#2

Hi,
Well can anyone tell me how to find the natural logarithm of a complex number p + iq.
Also please tell me how to convert it into logarithm to the base 10.
An external link to a webpage (where all the details are given) will be appreciated.

Tide Jun22-06 02:05 PM

Write the complex number in polar form and it should be evident how to find the logarithm.

To convert the logarithm to base 10 use log z = ln z/ln 10.

HallsofIvy Jun22-06 05:27 PM

Of course, since the argument of a complex number (the angle part of the polar form) can have any multiple of [itex]2\pi[/itex] added to it, the log function is, like most complex functions, multi-valued.

mathwonk Jun23-06 03:11 PM

the defn of ln(z) is the integral of the differential dw/w along a path from w = 1 to w = z, but not passing through w=0. the ambiguity is in the choice of that path.

if you transform it to r,theta coordinates I am guessing that differential becomes dr/r + i dtheta. since dr/r has a primitive, namely it equals dln|r|, which is dwefined everywhere except at 0, that part of the integral is not dependent on the path.

but the other part idtheta, does depend on how much angle is swept out wrt the origin by the whole path.


so the real part of the log of z is just the log of the absolute value |z|, but the complex part equals i times some determination of the angle arg(z).

I didn't really calculate this out just now, but this is probably very close to correct, as this is one of my long time favorite objects. I never understood complex functions or logs until I saw it explained roughly this way in Courant's calculus book (where else?).

mathwonk Jun23-06 03:13 PM

thats why 3^(ipi) = -1, i.e. minus one has absolute value 1 so real log zero, but angle pi, so complex part ipi.

and also why e^(2ipi) = 1, since one determination of the angle of 1 is 2pi.

Kaan Jan1-08 08:29 AM

I remember something like

Ln(z) = Ln(|z|) + iArg(z)

where |z| is the amplitude
and arg(z) is the angle in radians between the point z and x axis.

HallsofIvy Jan1-08 12:03 PM

Quote:

Quote by Kaan (Post 1557061)
I remember something like

Ln(z) = Ln(|z|) + iArg(z)

where |z| is the amplitude
and arg(z) is the angle in radians between the point z and x axis.

And, again, since the angle Arg(z) + [itex]2kpi[/itex], for any integer i, gives the same point, ln(z)= ln(|z|)+ i Arg(z)+ [itex]2ki\pi[/itex]

rbj Jan1-08 09:48 PM

Quote:

Quote by mathwonk (Post 1018788)
the defn of ln(z) is the integral of the differential dw/w along a path from w = 1 to w = z, but not passing through w=0. the ambiguity is in the choice of that path.

i don't really want to start anything, wonk, but while i believe that is true, is it the definition?

i didn't even think that was the definition for the real logarithm

[tex] \log(x) \equiv \int_1^x \frac{1}{u} du [/tex]

i thought that the definition had more to do with

[tex] \log(x y) = \log(x) + \log(y) [/tex]

that's what defines some function as a logarithm.

then it can be shown that for the function:

[tex] f_A(x) \equiv \int_1^x \frac{A}{u} du [/tex]

[tex] f_A(xy) = \int_1^{xy} \frac{A}{u} du = \int_1^{x} \frac{A}{u} du + \int_x^{xy} \frac{A}{u} du = \int_1^{x} \frac{A}{u} du + \int_1^{y} \frac{A}{u} du = f_A(x) + f_A(y)[/tex]

so we can say that fA is logarithmic. and the natural logarithm was such that numerator A is 1.

isn't this a difference between what is definition and what is a resulting property?

Vid Jan1-08 09:58 PM

They're both perfectly valid definitions. Some properties are easier to prove using one choice of definition over the other.
An example is this PDF using an alternate definition of the exp(x), and deriving some specific results from that.
http://www.math.lsu.edu/~mcgehee/Exp.pdf

HallsofIvy Jan2-08 04:50 AM

Quote:

Quote by rbj (Post 1557588)
i don't really want to start anything, wonk, but while i believe that is true, is it the definition?

i didn't even think that was the definition for the real logarithm

[tex] \log(x) \equiv \int_1^x \frac{1}{u} du [/tex]

i thought that the definition had more to do with

[tex] \log(x y) = \log(x) + \log(y) [/tex]

that's what defines some function as a logarithm.

then it can be shown that for the function:

[tex] f_A(x) \equiv \int_1^x \frac{A}{u} du [/tex]

[tex] f_A(xy) = \int_1^{xy} \frac{A}{u} du = \int_1^{x} \frac{A}{u} du + \int_x^{xy} \frac{A}{u} du = \int_1^{x} \frac{A}{u} du + \int_1^{y} \frac{A}{u} du = f_A(x) + f_A(y)[/tex]

so we can say that fA is logarithmic. and the natural logarithm was such that numerator A is 1.

isn't this a difference between what is definition and what is a resulting property?

What makes you think there is such a thing as such a thing as "the" definition of ln(x)?
Any function can have many different equivalent definitions. though it is a matter of taste, I would think that a definition that gives a direct formula ([itex]ln(x)= \int_1^x dt/t[/itex]) is better than a definition that says the fnction has such and such properties (ln(x) is the function f such that f(xy)= f(x)+ f(y) and f(1)= 0)/

rbj Jan2-08 01:32 PM

Quote:

Quote by HallsofIvy (Post 1557819)
What makes you think there is such a thing as such a thing as "the" definition of ln(x)?

i guess i normally think that a definition comes first, chronologically in the history or pedagogy.

in that sense, i think of logarithms as the generalized exponent of some given base (which is "e" if it's a "natural logarithm" - why e takes on the value that it does has to do with that integral, that wonk was using as a defining expression).

i didn't like it, but in my old calc book (Seeley, and i generally really like the book), they simple defined the natural exponential of a complex number as

[tex] e^z = e^{\mathrm{Re}(z)} \cos\left(\mathrm{Im}(z)\right) \ + \ i e^{\mathrm{Re}(z)} \sin\left(\mathrm{Im}(z)\right) [/tex]

and then stated that the definition was a good one since it reverted to the previous definition of ex for a real argument (when Im(z)=0) and satisfied ez+w = ez ew for any complex z and w. for me, that was an unsatisfying "definition".

BCox Jun29-09 11:06 PM

Re: To find logarithm of a complex number
 
Hello:

So if my function was

LN [ -exp(cx)/sin(d) - i ] = ?

Then, I would decompose it to real and imaginary parts?

The larger problem is

LN [ -exp(cx)/sin(d) - i ] - LN [ -exp(cx)/sin(d) + i ]

where c and d are constants.

???

HallsofIvy Jun30-09 05:53 AM

Re: To find logarithm of a complex number
 
Quote:

Quote by BCox (Post 2254916)
Hello:

So if my function was

LN [ -exp(cx)/sin(d) - i ] = ?

Then, I would decompose it to real and imaginary parts?

Assuming that -exp(cx)/sin(d) is real, that is of the form ln(a- i) with a= -exp(cx)/sin(d). Rewrite a- i in "polar form", [itex]re^{i\theta}[/itex] with [itex]r= \sqrt{exp(2cx)/sin^2(d)+ 1}[/itex][itex]= \sqrt{exp(2cx)+ sin^2(d)}/sin(d)[/itex] and [itex]\theta= arctan(sin(d)/exp(cx))[/itex]. Then [itex]ln(a- i)= ln(r)+ \theta i+ 2k\pi i[/itex] or
[tex]ln(-exp(cx)/sin(d)- i)= \frac{1}{2}ln(exp(2cx)+ sin^2(d))- ln(sin(d))+ arctan(sin(d)/exp(cx))i+ 2k\pi i[/tex]

Quote:

The larger problem is

LN [ -exp(cx)/sin(d) - i ] - LN [ -exp(cx)/sin(d) + i ]

where c and d are constants.

???

BCox Jun30-09 09:33 PM

Re: To find logarithm of a complex number
 
Hmm... perhaps a rephrase of the question is more appropriate.

I want to take the limit of the following function as x -> infinity

(1/d) * arcot [ - exp(cx) / sin(d) ]

where d is (-pi,0)

Now a prior condition would stipulate that the above has to go to zero as x tends to infinity. But I am more interested at the rate at which the function goes to zero as x -> infinity.

For one case as d->0, the first term tends to infinity. But I have another condition telling me that the whole function must approach zero; so that the second term arcot() must approach zero at a faster rate than 1/d. But analytically, what is the rate that

(1/d) * arcot [ - exp(cx) / sin(d) ] ~ ? as x->infinity

clustro Jun30-09 11:15 PM

Re: To find logarithm of a complex number
 
I'm not really a "mathematician", but for what its worth, you could put the complex number into a power series and hopefully try to see what it converges too...no? :( ??

HallsofIvy Jul1-09 05:24 AM

Re: To find logarithm of a complex number
 
?? A power series is necessarily a function. How do you put a number in a power series.

BCox Jul1-09 07:22 AM

Re: To find logarithm of a complex number
 
Quote:

Quote by HallsofIvy (Post 2256291)
?? A power series is necessarily a function. How do you put a number in a power series.

right. It's tricky

BCox Jul1-09 07:57 AM

Re: To find logarithm of a complex number
 
Perhaps this may be correct....

1. arcot = ( pi/2 - arctan() )
2. lim 1/d * [ pi/2 - arctan() ] as x->infin
3. We know that the express above goes to zero. We just want to know how fast. So
lim 1/d * [ pi/2 - arctan() ] = 0 as x->infin
4. 1/d * arctan[ - exp(cx) / sin(d) ] = pi/2d
5. arctan[ - exp(cx) / sin(d) ] = pi/2
6. tan{ arctan[ - exp(cx) / sin(d) ] } = tan(pi/2)
7. - exp(cx) / sin(d) = infin

8. So that the rate at which
lim 1/d * [ pi/2 - arctan() ] -> 0 as x->infin
is equal to the rate at which
lim -exp(cx) / sin(d) -> infinity as x->infin

9. Now if we are correct to assume that
lim exp(x) -> infinity as x->infinity
is at the same rate as
lim exp(-x) -> 0 as x-> infinity

10. Then the rate at which lim -exp(cx) / sin(d) -> infinity as x->infin
is the the same as
lim sin(d) * exp(-cx) -> 0 as x->infin

11. And by communicative property we would have that
lim 1/d * [ pi/2 - arctan() ] -> 0 as x->infin
is the same as
lim sin(d) * exp(-cx) -> 0 as x->infin
Or the rae at which the original expression approaches zero is
sin(d) * exp(-cx)


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