HELP! Does i = 0 in complex numbers? (1 Viewer)

[hyparzero]

I was doing some questions on euler's formula when i noticed the following


e^iπ = - 1

=> e^2iπ = (- 1)² .... (squaring both sides)
=> e^2iπ = 1

=> ln [ e^2iπ ] = ln(1) .... (taking natural logs of both sides)

.'. 2iπ = 0 ... (since ln(1) = 0 )

Now, how can 2iπ = 0?
Since both i and Pi are a constants....<img>

I have looked over it for ages and still cant figure it out... care to help?

 

[Templar]

Check the thread in the extracurricular section on complex numbers, in particular log and principle log.

 

[hyparzero]

The first thing that popped into my head when I saw this topic was that you cannot perform normal operations on the complex plane.

Why not? I mean, Euler's formula can be proved using normal operations:

Let z = cosθ + isinθ

.'. dz/dθ = -sinθ + icosθ
=> dz/dθ = i( isinθ + cosθ )
=> dz/dθ = i( cosθ + isinθ )
=> dz/dθ = iz

.'. ∫1/z dz = i∫dθ

=> ln(z) = iθ

.'. z = e^iθ

If θ = π, then z = -1 ........ (since cos(π) + isin(π) = -1)

.'. e^iπ = - 1

I see normal operations here.... why doesnt it apply for the special case of Euler.

 

[hyparzero]

ah... i knew something was up..

just in case anyone cares..
e^2iπ = 1

(after taking natural logs of both sides, we have (correctly) ):

ln [ e^2iπ ] = ln(1)+ i(Arg(1)+2*Pi*k)

where k is any integer

yep, that solves my problem, thanks

 

[hyparzero]

The first line, you've dealt with it in the complex field.

e^reals > 0

Squaring it; you take 2ipi into real plane, and thus the real part = 0, whereas, you've ignored the imaginary part.

hmmm... it seems to make sense, but isnt 2iPi the imaginary part?

 

[SeDaTeD]

The regular log "function" over the complex numbers is actually a multivalued function: it has an infinite no. of values differing by multiples of 2pi*i.

log(z) = log|z| + i*arg(z). Note that arg(z) is a multivalued function.

The principle Log is analytic (*) on the complex cut plane ( C without the real interval (-inf,0] ).

Log(z) = log|z| + i*Arg(z), where Arg(z) takes values in (-pi,pi]. (This is called the principle arguement).

Note that you can't work with Log of negative real numbers, as it is not defined there, thus it does not make sense to talk about Log(-1).

(*) Analytic basically means that there is a power series representation of the function.

This is a bit similar to say sin and its inverse arcsin. arcsin would be multivalued if you didn't restrict it to only take values between -pi and pi. In the same sense, if you look at the complex exp function, its inverse log would be multivalued, thus the principle logarithm is defined to only have its imaginary part between -pi and pi. If you are just working in the real numbers, the regular log would be fine.

 

[Slidey]

In z=e^(2ipi+2ikpi)=e^(2ipi(1+k)), let k=-1.