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Homework Statement
Prove ei[itex]\theta[/itex] = cos(θ) + isin(θ)
Homework Equations
The Attempt at a Solution
Let g = cos(θ) + isin(θ)
[itex]\frac{dg}{dθ}[/itex] = -sin(θ) + icos(θ)
=> [itex]\frac{dg}{dθ}[/itex] = ig
=> [itex]\frac{dθ}{dg}[/itex] = [itex]\frac{1}{ig}[/itex]
=> [itex]\frac{dθ}{dg}[/itex] = -i[itex]\times[/itex][itex]\int[/itex][itex]\frac{1}{g}[/itex]
=> θ = -i[itex]\times[/itex]ln|g| + c
=> iθ - ic = ln|g|
=> |g| = e[itex]^{iθ - ic}[/itex]
g(0)=1, => c = 0
When g=1,θ=0
=> |g| = e[itex]^{iθ}[/itex]
Here's where the problem arises. Wouldn't that mean that g = [itex]\pm[/itex]e[itex]^{iθ}[/itex]
Help is appreciated. Thanks.
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