Question

How to take the log of negative and complex numbers.

Answer 1

blasphemy!

Answer 2

ln(-e)

remember:
\[e^{i \pi}=-1\]
so we have\[\ln(e^{i \pi}e)=\ln(e^{i \pi})+\ln(e)=i \pi+1\]

Answer 3

k, done

Answer 4

lmao

Answer 5

but remember 1=(-1)^2 so

(e^(i*pi))^2=1

e^(i 2pi)=1

Yay.

Answer 6

don't forget the argument function.

Answer 7

I don't like argumentative functions, I prefer peaceful,wise functions.

Answer 8

Oh yeah and:

\[e^x=e^{i(-ix)}=\cos(-ix)+isin(-ix)=\cos(ix)-isin(ix)=e^x\]

Answer 9

Logarithms of Negative and Complex numbers:

Briefly, an imaginary unit i = √(-1) and i² = -1. In engineering, often j is used instead of i. A complex number is given by
z = a + bi, where a is the real part and bi is the imaginary part. If
a = 0, then z is called a pure imaginary number. If b = 0 then z is called a real number. Consequently, all real numbers are also complex numbers but not all complex numbers are in the set of reals. For example, -3 is a real and a complex number but
-3 + 2i is only a complex number. An excellent example to aide in the comprehension of logarithms of imaginary and complex numbers is Euler's identity. A beautiful formula like E = mc². Euler's identity states that

eˆ(ix) ≡ cos (x) + i ∙ sin (x)

where, as mentioned above, i = √(-1) and x represents any real number.

Suppose x = π/2. Then

eˆ(πi/2) = cos (π/2) + i ∙ sin (π/2)

eˆ(πi/2) = i [∵ cos (π/2) = 0 and sin (π/2) = 1]. Thus

ln [eˆ(πi/2)] = ln (i)

Applying the properties, we obtain

(πi/2)(ln e) = ln (i)

∴ ln (i) = πi/2 [∵ ln e = 1]


Similarly, if x = π/4, then

eˆ(πi/4) = cos (π/4) + i ∙ sin (π/4)

eˆ(πi/4) = (1/√2) + (1/√2)i [∵ cos (π/4) = sin (π/4) = 1/√2].

Thus

ln [eˆ(πi/4)] = ln [(1/√2) + (1/√2)i]

Applying the properties as above,

πi/4 = ln [(1/√2) + (1/√2)i]

∴ ln [(1/√2) + (1/√2)i] = πi/4


If x = π, then

eˆ(iπ) = cos (π) + i ∙ sin (π)

eˆ(iπ) = -1 [∵ cos (π) = -1 and sin (π) = 0]. Thus

ln [eˆ(iπ)] = ln (-1)

∴ ln (-1) = πi


If x = 0, then

eˆ(0i) = cos (0) + i ∙ sin (0)

eˆ(0i) = 1 [∵ cos (0) = 1 and sin (0) = 0]. Thus

ln [eˆ(0i)] = ln (1)

(0i)(ln e) = ln (1)

∴ ln (1) = 0 [∵ ln 1 = log 1 to base e and eˆ0 = 1]

Please note that since sin (x) and cos (x) are periodic functions, eˆ(iθ) = eˆ[(θ + 2πk)i], where θ is the angle measured in radians and k is any integer.

Answer 10

Don't forget your towel and hyperbolic functions, x=cosh(t) and y=sinh(t) which tell you where you are on the unit hyperbola, which obeys x^2-y^2=1.

\[\frac{ d^2 }{ dt^2 }(\cosh(t))=\cosh(t)\]

Yeah, not negative. Woah.

Answer 11

@calculusfunctions There are actually infinitely many complex logarithms to any number. Hooray.

Answer 12

Yes @Kainui I know. Thank you for the lesson. I'm a teacher and the above is short excerpt of a lesson I had written on "LOGARITHMS" for my class.

Answer 13

Ahhh cool.