- #1

- 58

- 0

dz = -sin(x) + icos(x)dx

= i(isin(x) + cos(x))dx

∫ dz/z = ∫ idx

ln(z) = ix

e^(ix) = z

e^(ix) = cos(x) + isin(x)

- Thread starter inknit
- Start date

- #1

- 58

- 0

dz = -sin(x) + icos(x)dx

= i(isin(x) + cos(x))dx

∫ dz/z = ∫ idx

ln(z) = ix

e^(ix) = z

e^(ix) = cos(x) + isin(x)

- #2

- 121

- 0

I think essentially you've shown some intuition for why it's a true formula: both z = cos(x) + i·sin(x) and z = e^{i·x} solve the ODE dz/dx = i·z, and both are 1 when x = 0. By uniqueness of solutions to ODEs, we should have e^{i·x} = cos(x) + i·sin(x).

Last edited:

- #3

- 166

- 0

Yes, but you have to make sure you have a good definition of the complex logarithm.

- #4

tiny-tim

Science Advisor

Homework Helper

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hi inknit!

ln(z) = ix + C …

*you still need to prove what C is! *

nooo …∫ dz/z = ∫ idx

ln(z) = ix

ln(z) = ix + C …

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