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Question Number 17713 by tawa tawa last updated on 09/Jul/17

$$\mathrm{Evaluate}:{(-\sqrt{3})}^{(-\sqrt{2})}$$

Commented by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17

$$a={(-\sqrt{3})}^{(-\sqrt{2})}$$$$lna=-\sqrt{2}ln(-\sqrt{3})=-\sqrt{2}ln\left(i^2\sqrt{3}\right)=$$$$=-\sqrt{2}[2lni+\frac{1}{2}ln3]=-\sqrt{2}[2i\frac{\pi }{2}+\frac{1}{2}ln3]=$$$$=-\frac{\sqrt{2}}{2}(2i\pi +ln3)$$$$\Rightarrow a=e^{-\frac{\sqrt{2}}{2}(2i\pi +ln3)}={\left(e^{2i\pi +ln3}\right)}^{\frac{-\sqrt{2}}{2}}=$$$$={[{\left(e^{i\pi }\right)}^2.e^{ln3}]}^{-\frac{\sqrt{2}}{2}}={[{(-1)}^2.3]}^{-\frac{\sqrt{2}}{2}}=3^{(-\frac{\sqrt{2}}{2})}$$$$note:$$$$1)i=cos90+isin90=e^{i\frac{\pi }{2}}\Rightarrow lni=i.\frac{\pi }{2}$$$$2)e^{i\pi }+1=0\left({Euler}^{\prime }sformula\right)$$

Commented by tawa tawa last updated on 09/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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