Question Number 197184 by pete last updated on 10/Sep/23
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{log}\left(−\mathrm{log}{i}\right)=\mathrm{log}\left(\frac{\pi}{\mathrm{2}}\right)−{i}\frac{\pi}{\mathrm{2}} \\ $$
Answered by Frix last updated on 10/Sep/23
$$\mathrm{ln}\:\left(−\mathrm{ln}\:\mathrm{i}\right)\:=\mathrm{ln}\:\frac{\pi}{\mathrm{2}}\:−\mathrm{i}\frac{\pi}{\mathrm{2}} \\ $$$$−\mathrm{ln}\:\mathrm{i}\:=\mathrm{e}^{\mathrm{ln}\:\frac{\pi}{\mathrm{2}}\:−\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$−\mathrm{ln}\:\mathrm{i}\:=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$−\mathrm{ln}\:\mathrm{i}\:=−\frac{\pi}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{ln}\:\mathrm{i}\:=\frac{\pi}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{i}=\mathrm{i}\:\mathrm{true} \\ $$
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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