Question
$\ln (\ln (\cos \mathrm{e}+\mathrm{i} \sin \mathrm{e}))$ is equal to(a) e(b) $1+\frac{\pi}{2} \mathrm{i}$(c) $\mathrm{e} \times \frac{\pi}{2} \times \mathrm{i}$(d) $\mathrm{e}+\frac{\pi}{2} \mathrm{i}$
Step 1
We know that $\cos \mathrm{e}+\mathrm{i} \sin \mathrm{e}$ is equal to $e^{i\mathrm{e}}$ by Euler's formula. So we can rewrite the expression as $\ln (\ln (e^{i\mathrm{e}}))$. Show more…
Show all steps
Your feedback will help us improve your experience
Uma Kumari and 96 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$$ \int\left(\frac{\ln x-1}{(\ln x)^{2}+1}\right)^{2} d x \text { is equal to } $$ (A) $\frac{x}{x^{2}+1}+c$ (B) $\frac{\ln x}{(\ln x)^{2}+1}+c$ (C) $\frac{x}{(\ln x)^{2}+1}+c$ (D) $e^{x}\left(\frac{x}{x^{2}+1}\right)+c$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD