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To find logarithm of a complex number 
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#1
Jun2206, 12:33 PM

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#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. 


#2
Jun2206, 02:05 PM

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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. 


#3
Jun2206, 05:27 PM

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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, multivalued.



#4
Jun2306, 03:11 PM

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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 dlnr, 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?). 


#5
Jun2306, 03:13 PM

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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. 


#6
Jan108, 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. 


#7
Jan108, 12:03 PM

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#8
Jan108, 09:48 PM

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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 f_{A} 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? 


#9
Jan108, 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 


#10
Jan208, 04:50 AM

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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)/ 


#11
Jan208, 01:32 PM

P: 2,251

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 e^{x} for a real argument (when Im(z)=0) and satisfied e^{z+w} = e^{z} e^{w} for any complex z and w. for me, that was an unsatisfying "definition". 


#12
Jun2909, 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. ??? 


#13
Jun3009, 05:53 AM

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[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] 


#14
Jun3009, 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 


#15
Jun3009, 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? :( ??



#16
Jul109, 05:24 AM

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?? A power series is necessarily a function. How do you put a number in a power series.



#17
Jul109, 07:22 AM

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#18
Jul109, 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) 


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