Question

$\log (\log \mathrm{i})$ equals

• A. $\log \frac{\mathrm{\pi }}{2}$
• B. $\log \mathrm{i}\frac{\mathrm{\pi }}{2}$
• C. $\log \frac{\mathrm{\pi }}{2}+\mathrm{i}\frac{\mathrm{\pi }}{2}$
• D. $\log \frac{\mathrm{\pi }}{2}-\mathrm{i}\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is C

$\log \frac{\mathrm{\pi }}{2}+\mathrm{i}\frac{\mathrm{\pi }}{2}$

Explanation of the correct option.

Compute the required value.

We know that, $\mathrm{i}={\mathrm{e}}^{\mathrm{i\theta }}$

${\mathrm{e}}^{\mathrm{i\theta }}$ can be written as $\cos \left(\mathrm{\theta }\right)+\mathrm{i}\sin \left(\mathrm{\theta }\right)$

$\mathrm{i}=$$\cos \left(\mathrm{\theta }\right)+\mathrm{i}\sin \left(\mathrm{\theta }\right)$

Since, the angle $\theta =\frac{\mathrm{\pi }}{2}$.

$\mathrm{i}={\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}$

Given: $\log (\log \mathrm{i})$

Substitute the value of $\mathrm{i}$.

$$\begin{gathered} \log (\log \mathrm{i})=\log (\log {\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}})\left[\because \log \left({\mathrm{a}}^{\mathrm{x}}\right)=\mathrm{x}\log \left(\mathrm{a}\right)\right] \\ \Rightarrow \log (\log \mathrm{i})=\log \left(\mathrm{i}\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \log (\log \mathrm{i})=\log \left(\frac{\mathrm{\pi }}{2}{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}\right)\left[\because \mathrm{i}={\mathrm{e}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}\right] \\ \Rightarrow \log (\log \mathrm{i})=\log \frac{\mathrm{\pi }}{2}+{\mathrm{loge}}^{\mathrm{i}\frac{\mathrm{\pi }}{2}}\left[\because \log \left(\mathrm{ab}\right)=\log \left(\mathrm{a}\right)+\log \left(\mathrm{b}\right)\right] \\ \Rightarrow \log (\log \mathrm{i})=\log \frac{\mathrm{\pi }}{2}+\mathrm{i}\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, option $\left(\mathrm{C}\right)$ is the correct answer.