Log of i

  • #1
184
0
Can you get logarithms of imaginary numbers?

The reason I was wondering is because integrating 1/(1+x^2) WRT x.
I know the answer is arctan(x), but how about breaking it into partial fractions by doing 1/(1-ix)(1+ix)?
 

Answers and Replies

  • #2
KTC
90
0
Yes, though it's a mutivalued function so it's not as simple as with natural number.
 
  • #3
quasar987
Science Advisor
Homework Helper
Gold Member
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Let's try to find log(i), given that log is the inverse function of the exponential defined on the whole complex plane.

So, log(i), whatever it is, should be a number such that exp(log(i))=i. Let's say log(i)=z=x+iy... then z is such that exp(z)=exp(x+iy)=exp(x)exp(iy)=exp(x)[cos(y)+isin(y)]=i. Well, x=0, y=pi/2 does the trick. But so does x=0, y=pi/2+2npi for any integer n. So log(i) is actually a multivalued function.
 
  • #4
Alkatran
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Given a complex number x = a + bi we want to find its logarithm
[tex]c + di = ln(a + bi)[/tex]

Take the exponential of both sides
[tex]e^{c + di} = e^{ln(a + bi)}[/tex]
[tex]e^c e^{di} = a + bi[/tex]
[tex]e^c (cos(\frac{d}{\pi}) + i sin(\frac{d}{\pi})) = a + bi[/tex]

Separating the imaginary and real terms:
[tex]e^c cos(\frac{d}{\pi}) = a[/tex]
[tex]e^c sin(\frac{d}{\pi}) = b[/tex]

To get d, divide one by the other:
[tex]\frac{e^c}{e^c} \frac{sin(\frac{d}{\pi})}{cos(\frac{d}{\pi})} = \frac{b}{a}[/tex]
[tex]tan(\frac{d}{\pi})} = \frac{b}{a}[/tex]
[tex]d = tan^{-1}(\frac{b}{a}) \pi[/tex]

To get c, use the absolute values and Pythagoras:
[tex]e^c e^{di} = a + bi[/tex]
[tex]|e^c e^{di}|^2 = |a + bi|^2[/tex]
[tex]|e^c|^2 = a^2 + b^2[/tex]
[tex]e^c = \sqrt{a^2 + b^2}[/tex]
[tex]c = ln(\sqrt{a^2 + b^2})[/tex]

Note that you can add/subtract 2pi to d to get more solutions, which is a typical side effect of using arctan.
 
Last edited:
  • #5
I haven't read everything you wrote jet...

but I had an Problem too:

i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
...

It's going on like that,
but if you make de lg of the rationaly numbers...

lg(i^0)=lg(1) ____ lg(i^2)=lg(-1) ____ lg(i^4)=lg(1) ...

then there will be...

0*lg(i)=lg(1) ____ 2*lg(i)=lg(-1) _____ 4*lg(i)=lg(1) ...

____ 0=lg(1) ______ lg(i)=lg(-1)/2 _____ lg(i)=lg(1)/4 ...

NOW IS THAT POSSIBLE???

lg(1)=0 ... That's True...
BUT:
lg(-1)=????

My first Problem...
Just forget it for the moment...

SO: lg(i)=0/4 => lg(i)=0 => lg(i)=lg(1) => i=1 !!!!!!!!

Now where was I wrong...???
or is it just that you can't work that way...


as an other example...

i^2= -1
i = -/-1]


i*i = -/ -1 ]*-/ -1 ]

So...

i*i = -/ -1*-1 ]
i*i = -/ 1 ]
i*i = 1 ???????

Both time's i=1...
 
  • #6
353
1
I haven't read everything you wrote jet...

but I had an Problem too:

i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
...

It's going on like that,
but if you make de lg of the rationaly numbers...
lg(i^0)=lg(1) ____ lg(i^2)=lg(-1) ____ lg(i^4)=lg(1) ...

then there will be...

0*lg(i)=lg(1) ____ 2*lg(i)=lg(-1) _____ 4*lg(i)=lg(1) ...

____ 0=lg(1) ______ lg(i)=lg(-1)/2 _____ lg(i)=lg(1)/4 ...

NOW IS THAT POSSIBLE???

lg(1)=0 ... That's True...
BUT:
lg(-1)=???? -> log(-1)=[π/ln(10)]i and in case your wondering ln(-1)=πi

My first Problem...
Just forget it for the moment...

SO: lg(i)=0/4 => lg(i)=0 => lg(i)=lg(1) => i=1 !!!!!!!!

Now where was I wrong...???
or is it just that you can't work that way...


as an other example...

i^2= -1
i = -/-1]


i*i = -/ -1 ]*-/ -1 ]

So...

i*i = -/ -1*-1 ]
i*i = -/ 1 ]
i*i = 1 ???????

Both time's i=1...

The fallacy is that √(xy)=√x√y is generally valid only if atleast one of the two numbers x or y is positive. (From Wiki)
 
  • #7
2,981
5
[tex]
\log{(i)} = i \, \left(\frac{\pi}{2} + 2 \, k \, \pi\right), \; k \in \mathbb{Z}
[/tex]
 
  • #8
329
0
Can you get logarithms of imaginary numbers?

The reason I was wondering is because integrating 1/(1+x^2) WRT x.
I know the answer is arctan(x), but how about breaking it into partial fractions by doing 1/(1-ix)(1+ix)?
You don't actually need to find ln(i) in order to perform the integration using partial fractions, but I think you will need the following formula:

[tex]\tan^{-1}(z) = \frac{1}{2i} \ln \frac{1-iz}{1+iz}[/tex]

You might want to see if you can derive it (it's not hard).
 
  • #9
log (i) = ?

log (i) = x

convert imaginary to exponential form...

thus;

(r)(exp^i) => i = (1)exp^(∏/2)i

then substitute to equation...

log (1exp^(∏/2)i)

basic logarithmic law...

log (ab) = log a + log b

therefore...

log (1exp^(∏/2)i) = log 1 + log exp^((∏/2)i)

log 1 = 0

log exp^((∏/2)i) = i log exp^(∏/2) = 0.6822 i

therefore...

log (i) = 0.6822i
 
  • #10
another way to solve log (i)

log (i) = ?

let log (i) = x

therefore...

10^x = i

take the natural logarithm...

ln 10^x = ln i

ln 10^x = x ln 10

ln (i) = ln (1 exp^((∏/2)i) = ln 1 + ln exp^(∏/2)i = (∏/2)i

x ln 10 = (∏/2)i

x = (∏/2)i / ln 10

x = 0.6822 i

same as the answer of my first post...
different solution but one answer...
 
  • #11
pwsnafu
Science Advisor
1,080
85
same as the answer of my first post...
different solution but one answer...

If you want a unique answer you need to pick a branch cut first.
 
  • #12
132
4
There are logarithms of [itex]i[/itex], but in terms of the calculus, your problem asks you to differentiate with respect to a [itex]x \in \mathbb{R}[/itex], right?

While the complex numbers contains all the real numbers, you may need to justify why you can get a solution of a real variable using a method with complex numbers.
 

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