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Question Number 59833 by bhanukumarb2@gmail.com last updated on 15/May/19

Answered by tanmay last updated on 15/May/19

$$e^{i\theta }=cos\theta +isin\theta$$$$e^{-i\theta }=cos\theta -isin\theta$$$$cos\theta =\frac{e^{i\theta }+e^{-i\theta }}{2}\to cosi=\frac{e^{-1}+e^1}{2}=a$$$$isin\theta =\frac{e^{i\theta }-e^{-i\theta }}{2}\to isini=\frac{e^{-1}-e^1}{2}=b$$$$a>b$$$$coti=\frac{cosi}{sini}=\frac{\frac{e^{-1}+e^1}{2}}{\frac{e^{-1}-e^1}{2i}}=\frac{a}{\frac{b}{i}}=i\left(\frac{a}{b}\right)$$$$tani=\frac{sini}{cosi}=\frac{\frac{e^{-1}-e}{2i}}{\frac{e^{-1}+e}{2}}=\frac{\frac{b}{i}}{a}=\left(\frac{b}{a}\right)\times \frac{1}{i}=-\left(\frac{b}{a}\right)i$$$$\underset{t\to \mathrm{\infty }}{\lim }{\left(\frac{{\left(cosi\right)}^t+{\left(isini\right)}^t}{{\left(tani\right)}^t+{\left(coti\right)}^t}\right)}^{\frac{1}{t}}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[\frac{a^t+b^t}{{\left(\frac{-bi}{a}\right)}^t+{\left(\frac{a}{b}i\right)}^t}\right]}^{\frac{1}{t}}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[\frac{a^t+b^t}{\frac{{(-b)}^t\times i^t}{a^t}+\frac{a^t}{b^t}\times i^t}\right]}^{\frac{1}{t}}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[\frac{a^t{b}^t(a^t+b^t)}{b^{2t}\times {(-1)}^t\times i^t+a^{2t}\times i^t}\right]}^{\frac{1}{t}}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[\frac{1}{i^t}\times \frac{a^{2t}{b}^t+a^t{b}^{2t}}{a^{2t}+{(-1)}^t{b}^{2t}}\right]}^{\frac{1}{t}}\to [a>b]$$$$$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[\frac{1}{i^t}\times \frac{b^t+\frac{b^{2t}}{a^t}}{1+(-1)\frac{b^{2t}}{a^{2t}}}\right]}^{\frac{1}{t}}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[{\left(\frac{b}{i}\right)}^t\right]}^{\frac{1}{t}}$$$$=\frac{b}{i}=\frac{e^{-1}-e^1}{2i}=\frac{i(e-\frac{1}{e})}{2}=z$$$$imz=\frac{e-\frac{1}{e}}{2}$$$${\left(imz\right)}^2+1$$$$=\frac{e^2-2+\frac{1}{e^2}}{4}+1$$$$=\frac{e^2+2+\frac{1}{e^2}}{4}={\left(\frac{e+\frac{1}{e}}{2}\right)}^2$$$$so$$$$imz+\sqrt{{\left(imz\right)}^2+1}$$$$=\frac{e-\frac{1}{e}}{2}+\frac{e+\frac{1}{e}}{2}$$$$=eproved$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$y=\underset{t\to \mathrm{\infty }}{\mathrm{li}}$$$$$$

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