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Question Number 197184 by pete last updated on 10/Sep/23

$$\mathrm{Show}\mathrm{that}\log (-\log i)=\log \left(\frac{\pi }{2}\right)-i\frac{\pi }{2}$$

Answered by Frix last updated on 10/Sep/23

$$\ln (-\ln \mathrm{i})=\ln \frac{\pi }{2}-\mathrm{i}\frac{\pi }{2}$$$$-\ln \mathrm{i}={\mathrm{e}}^{\ln \frac{\pi }{2}-\mathrm{i}\frac{\pi }{2}}$$$$-\ln \mathrm{i}=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{2}}$$$$-\ln \mathrm{i}=-\frac{\pi }{2}\mathrm{i}$$$$\ln \mathrm{i}=\frac{\pi }{2}\mathrm{i}$$$$\mathrm{i}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}$$$$\mathrm{i}=\mathrm{i}\mathrm{true}$$

Commented by pete last updated on 10/Sep/23

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

Commented by Frix last updated on 10/Sep/23

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