Trigo, Log or Complex number?

Algebra Level 3

Find the value of sin ⁡ ( ln ⁡ ( i i ) ) \sin (\ln(i^{i}))


The answer is -1.

Harvindersingh Chavan by Harvindersingh Chavan , 24, India

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2 solutions

I think it is clearer this way.

sin ⁡ ( ln ⁡ ( i i ) ) = sin ⁡ ( i ln ⁡ ( i ) ) = sin ⁡ ( i ln ⁡ ( e π 2 i ) ) = sin ⁡ ( i 2 π 2 ) = sin ⁡ ( − π 2 ) = − sin ⁡ π 2 = − 1 \begin{aligned} \sin{\left(\ln{(i^i)}\right)} & = \sin{\left(i\ln{(i)}\right)} = \sin{\left(i\ln{(e^{\frac{\pi}{2}i})}\right)} = \sin{\left(i^2\frac{\pi}{2}\right)} \\ & = \sin{\left(-\frac{\pi}{2}\right)} = -\sin{\frac{\pi}{2}} = \boxed{-1} \end{aligned}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

sin ⁡ ( l n ( i i ) ) \sin (ln(i^{i})) = sin ⁡ ( l n ( e ( i π / 2 ) i ) \sin (ln(e^{(iπ/2)^{i}}) = sin ⁡ ( l n ( e − π / 2 ) \sin (ln(e^{-π/2}) = sin ⁡ ( − π / 2 ) \sin (-π/2) = -1

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Can you explain why i i = e − π / 2 i^i = e^{-\pi /2} ?

Yes challenge master

i = cos ⁡ π / 2 + i sin ⁡ π / 2 = e ( i π / 2 ) i = \cos π/2 + i \sin π/2 = e^{(iπ/2)}

Now, i i i^{i} = e ( i π / 2 ) i e^{(iπ/2)^{i}} and since i×i = -1 therefore i i = e − π / 2 i^i = e^{-\pi /2}

Harvindersingh Chavan - 6 years ago

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