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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

$$\sqrt{\mathrm{i}}=\sqrt{{\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}$$$${\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}\right)}^{\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}}=[{\left(a^b\right)}^c=a^{bc}]$$$$$$$$={\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i})}={\mathrm{e}}^{-\frac{\pi \sqrt{2}}{8}+\frac{\pi \sqrt{2}}{8}\mathrm{i}}=$$$$$$$$={\mathrm{e}}^{-\frac{\pi \sqrt{2}}{8}}{\mathrm{e}}^{\frac{\pi \sqrt{2}}{8}\mathrm{i}}=[r{\mathrm{e}}^{\mathrm{i}\theta }=r\cos \theta +\mathrm{i}r\sin \theta ]$$$$$$$$={\mathrm{e}}^{-\frac{\pi \sqrt{2}}{8}}\cos \frac{\pi \sqrt{2}}{8}+{\mathrm{ie}}^{-\frac{\pi \sqrt{2}}{8}}\sin \frac{\pi \sqrt{2}}{8}$$

Answered by Spillover last updated on 21/May/23

$$Lety={\left(\sqrt{i}\right)}^{\sqrt{i}}$$$$\ln y=\ln {\left(\sqrt{i}\right)}^{\sqrt{i}}$$$$\ln y=\sqrt{i}\ln \sqrt{i}$$$$e^{i\theta }=\cos \theta +i\sin \theta$$$$when\theta =\frac{\pi }{2}\to e^{i\frac{\pi }{2}}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i$$$$e^{i\frac{\pi }{2}}=i{\left(e^{i\frac{\pi }{2}}\right)}^{\frac{1}{2}}={\left(i\right)}^{\frac{1}{2}}=e^{i\frac{\pi }{4}}={\left(i\right)}^{\frac{1}{2}}=\sqrt{i}$$$$\ln y=\sqrt{i}\ln \sqrt{i}\ln y=e^{i\frac{\pi }{4}}\ln \left(e^{i\frac{\pi }{4}}\right)$$$$\ln y=e^{i\frac{\pi }{4}}\ln \left(e^{i\frac{\pi }{4}}\right)$$$$\ln y=e^{i\frac{\pi }{4}}.\frac{i\pi }{4}$$$$\ln y=(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}).\frac{i\pi }{4}$$$$\ln y=(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}).\frac{i\pi }{4}=(\frac{i\pi \sqrt{2}}{8}-\frac{\pi \sqrt{2}}{8})$$$$\ln y=(\frac{i\pi \sqrt{2}}{8}-\frac{\pi \sqrt{2}}{8})=(\frac{-\pi \sqrt{2}}{8}+\frac{i\pi \sqrt{2}}{8})$$$$e^{(\frac{-\pi \sqrt{2}}{8}+\frac{i\pi \sqrt{2}}{8})}=e^{-\frac{\pi \sqrt{2}}{8}}.e^{\frac{i\pi \sqrt{2}}{8}}$$$$e^{-\frac{\pi \sqrt{2}}{8}}.(\cos \frac{\pi \sqrt{2}}{8}+i\sin \frac{\pi \sqrt{2}}{8})$$$$(e^{-\frac{\pi \sqrt{2}}{8}}.\cos \frac{\pi \sqrt{2}}{8}+{ie}^{-\frac{\pi \sqrt{2}}{8}}\sin \frac{\pi \sqrt{2}}{8})$$$$realpart{e}^{-\frac{\pi \sqrt{2}}{8}}.\cos \frac{\pi \sqrt{2}}{8}$$$$Immarginarypart{e}^{-\frac{\pi \sqrt{2}}{8}}\sin \frac{\pi \sqrt{2}}{8}$$

Commented by Frix last updated on 21/May/23

$$\mathrm{You}\mathrm{need}17\mathrm{lines}\mathrm{where}\mathrm{I}\mathrm{need}5.\mathrm{Congrats}$$$${\mathrm{you}}^{\prime }\mathrm{re}240\mathrm{\%}\mathrm{ahead}!$$$$\mathit{B}\mathit{U}\mathit{T}:\mathrm{You}\mathrm{lost}\pi \mathrm{somewhere}.$$

Commented by Frix last updated on 21/May/23

$$\mathrm{See}\mathrm{question}192580$$

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