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Question-192542

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Question Number 192542 by peter frank last updated on 20/May/23
Answered by Frix last updated on 20/May/23
(√i)=(√e^(i(π/2)) )=e^(i(π/4)) =((√2)/2)+((√2)/2)i (e^(i(π/4)) )^(((√2)/2)+((√2)/2)i) = [(a^b )^c =a^(bc) ] =e^(i(π/4)(((√2)/2)+((√2)/2)i)) =e^(−((π(√2))/8)+((π(√2))/8)i) = =e^(−((π(√2))/8)) e^(((π(√2))/8)i) = [re^(iθ) =rcos θ +irsin θ] =e^(−((π(√2))/8)) cos ((π(√2))/8) +ie^(−((π(√2))/8)) sin ((π(√2))/8)
$$\sqrt{\mathrm{i}}=\sqrt{\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} }=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$$$\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}} =\:\:\:\:\:\:\:\:\:\:\left[\left({a}^{{b}} \right)^{{c}} ={a}^{{bc}} \right] \\ $$$$ \\ $$$$=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)} =\mathrm{e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}+\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{i}} = \\ $$$$ \\ $$$$=\mathrm{e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} \mathrm{e}^{\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{i}} =\:\:\:\:\:\:\:\:\:\:\left[{r}\mathrm{e}^{\mathrm{i}\theta} ={r}\mathrm{cos}\:\theta\:+\mathrm{i}{r}\mathrm{sin}\:\theta\right] \\ $$$$ \\ $$$$=\mathrm{e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} \mathrm{cos}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:+\mathrm{ie}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} \mathrm{sin}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$
Answered by Spillover last updated on 21/May/23
Let y=((√i) )^(√i) ln y=ln ((√i) )^(√i) ln y=(√i)ln (√i) e^(iθ) =cos θ+isin θ when θ=(π/2) → e^(i(π/2)) =cos (π/2)+isin (π/2)=i e^(i(π/2)) =i ( e^(i(π/2)) )^(1/2) =(i)^(1/2) = e^(i(π/4)) =(i)^(1/2) =(√i) ln y=(√i)ln (√i) ln y=e^(i(π/4)) ln (e^(i(π/4)) ) ln y=e^(i(π/4)) ln (e^(i(π/4)) ) ln y=e^(i(π/4)) .((iπ)/4) ln y=(cos (π/4)+isin (π/4)).((iπ)/4) ln y=(((√2)/2)+i((√2)/2)).((iπ)/4)=(((iπ(√2))/8)−((π(√2))/8)) ln y=(((iπ(√2))/8)−((π(√2))/8))=(((−π(√2))/8)+ ((iπ(√2))/8)) e^((((−π(√2))/8)+ ((iπ(√2))/8))) =e^(−((π(√2))/8)) .e^((iπ(√2))/8) e^(−((π(√2))/8)) .(cos ((π(√2))/8)+isin ((π(√2))/8)) (e^(−((π(√2))/8)) .cos ((π(√2))/8)+ie^(−((π(√2))/8)) sin ((π(√2))/8)) real part e^(−((π(√2))/8)) .cos ((π(√2))/8) Immarginary part e^(−((π(√2))/8)) sin ((π(√2))/8)
$${Let}\:\:\:{y}=\left(\sqrt{{i}}\:\right)^{\sqrt{{i}}} \: \\ $$$$\mathrm{ln}\:{y}=\mathrm{ln}\:\left(\sqrt{{i}}\:\right)^{\sqrt{{i}}} \:\: \\ $$$$\mathrm{ln}\:{y}=\sqrt{{i}}\mathrm{ln}\:\sqrt{{i}}\:\:\: \\ $$$${e}^{{i}\theta} =\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta \\ $$$${when}\:\:\:\:\theta=\frac{\pi}{\mathrm{2}}\:\:\rightarrow\:\:{e}^{{i}\frac{\pi}{\mathrm{2}}} =\mathrm{cos}\:\frac{\pi}{\mathrm{2}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{2}}={i} \\ $$$$\:{e}^{{i}\frac{\pi}{\mathrm{2}}} ={i}\:\:\:\left(\:{e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} =\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{{i}} \\ $$$$\mathrm{ln}\:{y}=\sqrt{{i}}\mathrm{ln}\:\sqrt{{i}}\:\:\:\:\:\:\:\mathrm{ln}\:{y}={e}^{{i}\frac{\pi}{\mathrm{4}}} \mathrm{ln}\:\left({e}^{{i}\frac{\pi}{\mathrm{4}}} \right) \\ $$$$\mathrm{ln}\:{y}={e}^{{i}\frac{\pi}{\mathrm{4}}} \mathrm{ln}\:\left({e}^{{i}\frac{\pi}{\mathrm{4}}} \right) \\ $$$$\mathrm{ln}\:{y}={e}^{{i}\frac{\pi}{\mathrm{4}}} .\frac{{i}\pi}{\mathrm{4}} \\ $$$$\mathrm{ln}\:{y}=\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right).\frac{{i}\pi}{\mathrm{4}} \\ $$$$\mathrm{ln}\:{y}=\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right).\frac{{i}\pi}{\mathrm{4}}=\left(\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{8}}−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\right) \\ $$$$\mathrm{ln}\:{y}=\left(\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{8}}−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\right)=\left(\frac{−\pi\sqrt{\mathrm{2}}}{\mathrm{8}}+\:\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\right) \\ $$$${e}^{\left(\frac{−\pi\sqrt{\mathrm{2}}}{\mathrm{8}}+\:\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\right)} ={e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} .{e}^{\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} \\ $$$${e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} .\left(\mathrm{cos}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}+{i}\mathrm{sin}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\right) \\ $$$$\left({e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} .\mathrm{cos}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}+{ie}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} \mathrm{sin}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\right) \\ $$$${real}\:{part}\:\:\:\:\:\:\:\:\:\:{e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} .\mathrm{cos}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$$${Immarginary}\:{part}\:\:{e}^{−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}} \mathrm{sin}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$
Commented by Frix last updated on 21/May/23
You need 17 lines where I need 5. Congrats you′re 240% ahead! BUT: You lost π somewhere.
$$\mathrm{You}\:\mathrm{need}\:\mathrm{17}\:\mathrm{lines}\:\mathrm{where}\:\mathrm{I}\:\mathrm{need}\:\mathrm{5}.\:\mathrm{Congrats} \\ $$$$\mathrm{you}'\mathrm{re}\:\mathrm{240\%}\:\mathrm{ahead}! \\ $$$$\boldsymbol{{BUT}}:\:\mathrm{You}\:\mathrm{lost}\:\pi\:\mathrm{somewhere}. \\ $$
Commented by Frix last updated on 21/May/23
See question 192580
$$\mathrm{See}\:\mathrm{question}\:\mathrm{192580} \\ $$

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