Question Number 59833 by bhanukumarb2@gmail.com last updated on 15/May/19
Answered by tanmay last updated on 15/May/19
![e^(iθ) =cosθ+isinθ e^(−iθ) =cosθ−isinθ cosθ=((e^(iθ) +e^(−iθ) )/2)→cosi=((e^(−1) +e^1 )/2)=a isinθ=((e^(iθ) −e^(−iθ) )/2)→isini=((e^(−1) −e^1 )/2)=b a>b coti=((cosi)/(sini))=(((e^(−1) +e^1 )/2)/((e^(−1) −e^1 )/(2i)))=(a/(b/i))=i((a/b)) tani=((sini)/(cosi))=(((e^(−1) −e)/(2i))/((e^(−1) +e)/2))=((b/i)/a)=((b/a))×(1/i)=−((b/a))i lim_(t→∞) ((((cosi)^t +(isini)^t )/((tani)^t +(coti)^t )))^(1/t) =lim_(t→∞) [((a^t +b^t )/((((−bi)/a))^t +((a/b)i)^t ))]^(1/t) =lim_(t→∞) [((a^t +b^t )/((((−b)^t ×i^t )/a^t )+(a^t /b^t )×i^t ))]^(1/t) =lim_(t→∞) [((a^t b^t (a^t +b^t ))/(b^(2t) ×(−1)^t ×i^t +a^(2t) ×i^t ))]^(1/t) =lim_(t→∞) [(1/i^t )×((a^(2t) b^t +a^t b^(2t) )/(a^(2t) +(−1)^t b^(2t) ))]^(1/t) →[a>b] =lim_(t→∞) [(1/i^t )×((b^t +(b^(2t) /a^t ))/(1+(−1)(b^(2t) /a^(2t) )))]^(1/t) =lim_(t→∞) [((b/i))^t ]^(1/t) =(b/i)=((e^(−1) −e^1 )/(2i))=((i(e−(1/e)))/2)=z imz=((e−(1/e))/2) (imz)^2 +1 =((e^2 −2+(1/e^2 ))/4)+1 =((e^2 +2+(1/e^2 ))/4)=(((e+(1/e))/2))^2 so imz+(√((imz)^2 +1)) =((e−(1/e))/2)+((e+(1/e))/2) =e proved y=li_(t→∞)](https://www.tinkutara.com/question/Q59875.png)
$${e}^{{i}\theta} ={cos}\theta+{isin}\theta \\ $$$${e}^{−{i}\theta} ={cos}\theta−{isin}\theta \\ $$$${cos}\theta=\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\rightarrow{cosi}=\frac{{e}^{−\mathrm{1}} +{e}^{\mathrm{1}} }{\mathrm{2}}={a} \\ $$$${isin}\theta=\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}}\rightarrow{isini}=\frac{{e}^{−\mathrm{1}} −{e}^{\mathrm{1}} }{\mathrm{2}}={b} \\ $$$${a}>{b} \\ $$$${coti}=\frac{{cosi}}{{sini}}=\frac{\frac{{e}^{−\mathrm{1}} +{e}^{\mathrm{1}} }{\mathrm{2}}}{\frac{{e}^{−\mathrm{1}} −{e}^{\mathrm{1}} }{\mathrm{2}{i}}}=\frac{{a}}{\frac{{b}}{{i}}}={i}\left(\frac{{a}}{{b}}\right) \\ $$$${tani}=\frac{{sini}}{{cosi}}=\frac{\frac{{e}^{−\mathrm{1}} −{e}}{\mathrm{2}{i}}}{\frac{{e}^{−\mathrm{1}} +{e}}{\mathrm{2}}}=\frac{\frac{{b}}{{i}}}{{a}}=\left(\frac{{b}}{{a}}\right)×\frac{\mathrm{1}}{{i}}=−\left(\frac{{b}}{{a}}\right){i} \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\left({cosi}\right)^{{t}} +\left({isini}\right)^{{t}} }{\left({tani}\right)^{{t}} +\left({coti}\right)^{{t}} }\right)^{\frac{\mathrm{1}}{{t}}} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{{a}^{{t}} +{b}^{{t}} }{\left(\frac{−{bi}}{{a}}\right)^{{t}} +\left(\frac{{a}}{{b}}{i}\right)^{{t}} }\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{a}^{{t}} +{b}^{{t}} }{\frac{\left(−{b}\right)^{{t}} ×{i}^{{t}} }{{a}^{{t}} }+\frac{{a}^{{t}} }{{b}^{{t}} }×{i}^{{t}} }\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{{a}^{{t}} {b}^{{t}} \left({a}^{{t}} +{b}^{{t}} \right)}{{b}^{\mathrm{2}{t}} ×\left(−\mathrm{1}\right)^{{t}} ×{i}^{{t}} +{a}^{\mathrm{2}{t}} ×{i}^{{t}} }\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{1}}{{i}^{{t}} }×\frac{{a}^{\mathrm{2}{t}} {b}^{{t}} +{a}^{{t}} {b}^{\mathrm{2}{t}} }{{a}^{\mathrm{2}{t}} +\left(−\mathrm{1}\right)^{{t}} {b}^{\mathrm{2}{t}} }\right]^{\frac{\mathrm{1}}{{t}}} \rightarrow\left[{a}>{b}\right] \\ $$$$ \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{{i}^{{t}} }×\frac{{b}^{{t}} +\frac{{b}^{\mathrm{2}{t}} }{{a}^{{t}} }}{\mathrm{1}+\left(−\mathrm{1}\right)\frac{{b}^{\mathrm{2}{t}} }{{a}^{\mathrm{2}{t}} }}\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left[\left(\frac{{b}}{{i}}\right)^{{t}} \right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$=\frac{{b}}{{i}}=\frac{{e}^{−\mathrm{1}} −{e}^{\mathrm{1}} }{\mathrm{2}{i}}=\frac{{i}\left({e}−\frac{\mathrm{1}}{{e}}\right)}{\mathrm{2}}={z} \\ $$$${imz}=\frac{{e}−\frac{\mathrm{1}}{{e}}}{\mathrm{2}} \\ $$$$\left({imz}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\frac{{e}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }}{\mathrm{4}}+\mathrm{1} \\ $$$$=\frac{{e}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }}{\mathrm{4}}=\left(\frac{{e}+\frac{\mathrm{1}}{{e}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${so}\: \\ $$$${imz}+\sqrt{\left({imz}\right)^{\mathrm{2}} +\mathrm{1}}\: \\ $$$$=\frac{{e}−\frac{\mathrm{1}}{{e}}}{\mathrm{2}}+\frac{{e}+\frac{\mathrm{1}}{{e}}}{\mathrm{2}} \\ $$$$={e}\:{proved} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${y}=\underset{{t}\rightarrow\infty} {\mathrm{li}} \\ $$$$ \\ $$