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Question Number 70138 by Scientist0000001 last updated on 01/Oct/19
prove that   e^(iθ) =e^(i(θ+2kΠ))   given that k=0,±1,±2...
$${prove}\:{that}\:\:\:{e}^{{i}\theta} ={e}^{{i}\left(\theta+\mathrm{2}{k}\Pi\right)} \:\:{given}\:{that}\:{k}=\mathrm{0},\pm\mathrm{1},\pm\mathrm{2}… \\ $$
Answered by mind is power last updated on 01/Oct/19
e^(ix) =cos(x)+isin(x)  ⇒e^(i(θ+2kπ)) =cos(θ+2kπ)+isin(2kπ+θ)  =cos(θ)+isin(θ)  we used periodicity of Sin and vos witch is 2π  cos(θ)+isin(θ)=e^(iθ)
$${e}^{{ix}} ={cos}\left({x}\right)+{isin}\left({x}\right) \\ $$$$\Rightarrow{e}^{{i}\left(\theta+\mathrm{2}{k}\pi\right)} ={cos}\left(\theta+\mathrm{2}{k}\pi\right)+{isin}\left(\mathrm{2}{k}\pi+\theta\right) \\ $$$$={cos}\left(\theta\right)+{isin}\left(\theta\right)\:\:{we}\:{used}\:{periodicity}\:{of}\:{Sin}\:{and}\:{vos}\:{witch}\:{is}\:\mathrm{2}\pi \\ $$$${cos}\left(\theta\right)+{isin}\left(\theta\right)={e}^{{i}\theta} \\ $$

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