To find logarithm of a complex number


by AlbertEinstein
Tags: complex, logarithm, number
AlbertEinstein
AlbertEinstein is offline
#1
Jun22-06, 12:33 PM
P: 113
#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.
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
Tide
Tide is offline
#2
Jun22-06, 02:05 PM
Sci Advisor
HW Helper
P: 3,149
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
HallsofIvy is offline
#3
Jun22-06, 05:27 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
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
mathwonk is offline
#4
Jun23-06, 03:11 PM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,421

To find logarithm of a complex number


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
mathwonk is offline
#5
Jun23-06, 03:13 PM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,421
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
Kaan is offline
#6
Jan1-08, 08:29 AM
P: 5
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
HallsofIvy is offline
#7
Jan1-08, 12:03 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
Quote Quote by Kaan View Post
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
rbj is offline
#8
Jan1-08, 09:48 PM
P: 2,265
Quote Quote by mathwonk View Post
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
Vid is offline
#9
Jan1-08, 09:58 PM
P: 420
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
HallsofIvy is offline
#10
Jan2-08, 04:50 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
Quote Quote by rbj View Post
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
rbj is offline
#11
Jan2-08, 01:32 PM
P: 2,265
Quote Quote by HallsofIvy View Post
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
BCox is offline
#12
Jun29-09, 11:06 PM
P: 16
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
HallsofIvy is offline
#13
Jun30-09, 05:53 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
Quote Quote by BCox View Post
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]

The larger problem is

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

where c and d are constants.

???
BCox
BCox is offline
#14
Jun30-09, 09:33 PM
P: 16
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
clustro is offline
#15
Jun30-09, 11:15 PM
P: 77
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
HallsofIvy is offline
#16
Jul1-09, 05:24 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
?? A power series is necessarily a function. How do you put a number in a power series.
BCox
BCox is offline
#17
Jul1-09, 07:22 AM
P: 16
Quote Quote by HallsofIvy View Post
?? A power series is necessarily a function. How do you put a number in a power series.
right. It's tricky
BCox
BCox is offline
#18
Jul1-09, 07:57 AM
P: 16
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)


Register to reply

Related Discussions
Complex number Calculus & Beyond Homework 12
complex number Calculus & Beyond Homework 0
2 Bits of help neede for study Logarithm+Complex number Precalculus Mathematics Homework 14
The complex logarithm Calculus & Beyond Homework 1
determination of logarithm (complex) Calculus 4