Log of i: Can you get logarithms of imaginary numbers?

[theperthvan]

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)?

 

[KTC]

Yes, though it's a mutivalued function so it's not as simple as with natural number.

 

[quasar987]

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.

 

[Alkatran]

Given a complex number x = a + bi we want to find its logarithm
$$c + di = ln(a + bi)$$

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

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

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

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

Note that you can add/subtract 2pi to d to get more solutions, which is a typical side effect of using arctan.

 

[MarkUS ViFe]

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...

 

[BloodyFrozen]

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)

 

[Dickfore]

$$\log{(i)} = i \, \left(\frac{\pi}{2} + 2 \, k \, \pi\right), \; k \in \mathbb{Z}$$

 

[awkward]

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:

$$\tan^{-1}(z) = \frac{1}{2i} \ln \frac{1-iz}{1+iz}$$

You might want to see if you can derive it (it's not hard).

 

[michael urbano]

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

 

[michael urbano]

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...

 

[pwsnafu]

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.

 

[thelema418]

There are logarithms of $i$, but in terms of the calculus, your problem asks you to differentiate with respect to a $x \in \mathbb{R}$, 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.