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Question Number 14148 by RasheedSindhi last updated on 28/May/17
Determine (a) i^i (b) ω^ω (c) i^ω (d) ω^i
$$\mathrm{Determine} \\ $$$$\left({a}\right)\:{i}^{{i}} \\ $$$$\left({b}\right)\:\omega^{\omega} \\ $$$$\left({c}\right)\:{i}^{\omega} \\ $$$$\left({d}\right)\:\omega^{{i}} \\ $$
Answered by prakash jain last updated on 28/May/17
a) i=e^(i(π/2)) i^i =(e^(i(π/2)) )^i =e^(−(π/2))
$$\left.{a}\right) \\ $$$${i}={e}^{{i}\frac{\pi}{\mathrm{2}}} \\ $$$${i}^{{i}} =\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \\ $$
Commented by RasheedSindhi last updated on 31/May/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by sma3l2996 last updated on 29/May/17
(b): ω^ω =(e^(i((2π)/3)) )^ω =e^(iω((2π)/3)) =e^(i((2π)/3)(−(1/2)+i((√3)/2))) =e^(−i((2π)/3)×(1/2)−((2π)/3)×((√3)/2)) =e^(−((√3)/3)π) ×e^(−i(π/3)) =e^((−π)/( (√3))) ×((1/2)−i((√3)/2))=−e^((−π)/( (√3))) ×(−(1/2)+i((√3)/2)) ω^ω =−e^((−π)/( (√3))) ×ω (a): i^i =(e^(i(π/2)) )^i =e^(−(π/2)) (c): i^ω =e^(iω(π/2)) =e^(i(π/2)(−(1/2)+i((√3)/2))) =e^(−((√3)/4)π−i(π/4)) i^ω =(e^(−((√3)/4)π) /( (√2)))(1−i) (d): ω^i =(e^(i((2π)/3)) )^i =e^(−((2π)/3))
$$\left({b}\right):\:\:\:\:\omega^{\omega} =\left({e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{\omega} ={e}^{{i}\omega\frac{\mathrm{2}\pi}{\mathrm{3}}} ={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} ={e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}\pi}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\pi} ×{e}^{−{i}\frac{\pi}{\mathrm{3}}} ={e}^{\frac{−\pi}{\:\sqrt{\mathrm{3}}}} ×\left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=−{e}^{\frac{−\pi}{\:\sqrt{\mathrm{3}}}} ×\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\omega^{\omega} =−{e}^{\frac{−\pi}{\:\sqrt{\mathrm{3}}}} ×\omega \\ $$$$\left({a}\right):\:\:{i}^{{i}} =\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \\ $$$$\left({c}\right):\:\:\:{i}^{\omega} ={e}^{{i}\omega\frac{\pi}{\mathrm{2}}} ={e}^{{i}\frac{\pi}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} ={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\pi−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${i}^{\omega} =\frac{{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\pi} }{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}−{i}\right) \\ $$$$\left({d}\right):\:\:\:\omega^{{i}} =\left({e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{{i}} ={e}^{−\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$
Commented by RasheedSindhi last updated on 31/May/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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