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Question-108678




Question Number 108678 by Rasikh last updated on 18/Aug/20
Answered by abdomsup last updated on 18/Aug/20
dont exist  let x_n =((1/n))^i   lim_(n→+∞) x_n   =o^i   but  x_n =e^(iln((1/n)) =e^(−iln(n))  =cos(lnn)+isin(lnn))   but cos(lnn) and sin(lnn) havent  any limit
$${dont}\:{exist}\:\:{let}\:{x}_{{n}} =\left(\frac{\mathrm{1}}{{n}}\right)^{{i}} \\ $$$${lim}_{{n}\rightarrow+\infty} {x}_{{n}} \:\:={o}^{{i}} \:\:{but} \\ $$$${x}_{{n}} ={e}^{{iln}\left(\frac{\mathrm{1}}{{n}}\right)\:={e}^{−{iln}\left({n}\right)} \:={cos}\left({lnn}\right)+{isin}\left({lnn}\right)} \\ $$$${but}\:{cos}\left({lnn}\right)\:{and}\:{sin}\left({lnn}\right)\:{havent} \\ $$$${any}\:{limit} \\ $$$$ \\ $$
Answered by Her_Majesty last updated on 18/Aug/20
z=re^(iθ) ∧r≥0∧θ∈R ⇒ z^i =r^i e^(−θ) =(1/e^θ )e^(ilnr)   this is defined for θ∈R∧r≠0  ⇒  0^i  is not defined
$${z}={re}^{{i}\theta} \wedge{r}\geqslant\mathrm{0}\wedge\theta\in\mathbb{R}\:\Rightarrow\:{z}^{{i}} ={r}^{{i}} {e}^{−\theta} =\frac{\mathrm{1}}{{e}^{\theta} }{e}^{{ilnr}} \\ $$$${this}\:{is}\:{defined}\:{for}\:\theta\in\mathbb{R}\wedge{r}\neq\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{0}^{{i}} \:{is}\:{not}\:{defined} \\ $$

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