proof-e-i-cos-isin-

Question Number 5153 by 1771727373 last updated on 23/Apr/16

p r o o f e^{i Θ} = c o s (Θ) + i s i n (Θ)

Answered by 123456 last updated on 23/Apr/16

lets f (θ) = \cos θ + i \sin θ, f (0) = 1

\frac{\partial f}{\partial θ} = - \sin θ + i \cos θ = i (\cos θ + i \sin θ) = i f

\frac{d f}{f} = i d θ

\ln f = i θ + k

f = e^{i θ + k} = e^{i θ} e^{k} = {c e}^{i θ}

f (0) = 1 \Rightarrow {c e}^{0} = c = 1

f (θ) = e^{i θ}

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