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