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Question Number 114857 by Rio Michael last updated on 21/Sep/20

$$\mathrm{we}\mathrm{know}\mathrm{that}$$$$e^{\pi i}=-1$$$$\Rightarrow \ln \left(e^{\pi i}\right)=\ln (-1)$$$$\pi i=\ln (-1).$$$$\mathrm{How}\mathrm{good}\mathrm{is}\mathrm{this}\mathrm{prove}?$$

Commented by malwan last updated on 21/Sep/20

$$isthatmean\pi =ln\left({}^{\sqrt{-1}}\sqrt{-1}\right)?$$

Commented by mr W last updated on 21/Sep/20

$$youcanevensay$$$$\pi =\ln \sqrt[i]{-1}$$

Commented by Rio Michael last updated on 21/Sep/20

$$\mathrm{really}\mathrm{sir}?$$

Commented by MJS_new last updated on 21/Sep/20

$${(-1)}^{\frac{1}{\mathrm{i}}}={(-1)}^{-\mathrm{i}}$$$$-1={\mathrm{e}}^{\mathrm{i}\pi }\Rightarrow {(-1)}^{-\mathrm{i}}={\mathrm{e}}^{\pi }$$

Commented by Rio Michael last updated on 21/Sep/20

$$\mathrm{thanks}\mathrm{prof}$$

Commented by Dwaipayan Shikari last updated on 21/Sep/20

$$log(-1)=log\left(e^{\pi i}\right)$$$$log\left(e^{2k\pi i+\pi i}\right)=log(-1)\Rightarrow log(-1)=\pi i(2k+1)$$$$i^{\frac{1}{i}}=e^{\frac{\pi }{2}}$$$$i^i=e^{-\frac{\pi }{2}}$$

Commented by MJS_new last updated on 21/Sep/20

$$\mathrm{yes}$$

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