Imagining the imaginary

Algebra Level 4

ln ( i i ) = k × π \large {\ln(i^i)} = k × \pi

Submit the value of k k .

Notation: i = 1 i = \sqrt{-1} denotes the imaginary unit .

(Consider the principal branch of the complex logarithm).


The answer is -0.5.

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3 solutions

Chew-Seong Cheong
Dec 22, 2016

Relevant wiki: Euler's Formula

ln ( i i ) = i ln i By Euler’s formula: e π 2 i = cos π 2 + i sin π 2 = i ln ( e π 2 i ) = i ( π 2 i ) = π 2 \begin{aligned} \ln(i^i) & = i \ln \color{#3D99F6} i & \small \color{#3D99F6} \text{By Euler's formula: } e^{\frac \pi 2 i} = \cos \frac \pi 2 + i \sin \frac \pi 2 \\ &= i \ln \left ( {\color{#3D99F6}e^{\frac \pi 2 i}} \right) \\ &= i \left (\frac \pi 2 i \right) \\ &= - \frac \pi 2 \end{aligned}

k = 0.5 \implies k = \boxed {-0.5}

Have you used e i θ = cos θ + i sin θ e^{i\theta} = \cos\theta + i\sin\theta ? ... Plugging θ = π 2 \theta =\frac{\pi}{2} , e i π 2 = I e^{\frac{i\pi}{2}} = I

Prokash Shakkhar - 4 years, 5 months ago

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Yes, I have done that. I will explain it.

Chew-Seong Cheong - 4 years, 5 months ago

why level 4 , it is very very easy ! no offence , they just giving rating to anything ! at most level 2 !

Dont worry about the level bro, it will automatically change as the number of people answer it. Next time please comment it under the solution, currently you have posted it as a solution. Thanks!

Ashish Menon - 4 years, 5 months ago

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oh ! sorry and thanks for telling ! :) i'll edit it sometime later !

A Former Brilliant Member - 4 years, 5 months ago

no no you shut up, he's right.

Rohit Nandwani - 1 year, 3 months ago
Ashish Menon
Dec 22, 2016

i i = ( e i × π 2 ) i = e π 2 ln ( i i ) = π 2 k = 0.5 \begin{aligned} \large{i^i} & = \large{{\left(e^{i × \frac{\pi}{2}}\right)}^{i}}\\ \\ & = \large{e^{-\frac{\pi}{2}}}\\ \\ \implies \large{\ln(i^i)} = \large{-\dfrac{\pi}{2}}\\ \\ \implies \large{k} = \large{\color{#69047E}{\boxed{-0.5}}} \end{aligned}

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