Log of i

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.

Homework Helper
Gold Member
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.

Homework Helper
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.

Last edited:
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...

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.