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## Homework Statement

Prove e

^{i[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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