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Calculus Level 2

What is Im \text{Im} i i i^i

i = 1 i=\sqrt{-1}

Im means imaginary part of the complex number


The answer is 0.

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3 solutions

Siddharth Bhatt
Mar 23, 2015

i i = e π 2 = 0.207 i^i = e^{\frac{-\pi}{2}} = 0.207 which is purely real and thus has no imaginary part so \text{which is purely real and thus has no imaginary part so} Im i i = 0 \text{Im}\ i^i = 0

@kritarth lohomi

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A Former Brilliant Member - 6 years, 2 months ago

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Thank you @Azhaghu Roopesh M

kritarth lohomi - 6 years, 2 months ago

Same from my side

siddharth bhatt - 6 years, 2 months ago

To be precise, i i i^i has no fixed real value. It has an infinite number of values that can be shown using Euler's formula. We get,

i i = e 2 π n π 2 , n Z \large i^i=e^{-2\pi n-\frac{\pi}{2}}~,~n\in\mathbb{Z}

As you can see, n n can take any integer value and we'll get different values. But the value mostly used is when n = 0 n=0 (according to standard conventions) and we get the answer as e π 2 0.20788 \large \boxed{e^{-\frac{\pi}{2}}\approx 0.20788} .

Prasun Biswas - 6 years, 2 months ago

how???????/?

Zeeshan Ali - 6 years, 2 months ago
Chew-Seong Cheong
Mar 23, 2015

i i = ( 0 + i ) i = ( cos π 2 + i sin π 2 ) i = ( e π 2 i ) i = e π 2 = e π 2 + 0 i i^i = (0+i)^i = (\cos {\frac{\pi}{2}} + i \sin {\frac{\pi}{2}})^i = ( e^{\frac{\pi}{2}i} )^i = e^{-\frac{\pi}{2}} = e^{-\frac{\pi}{2}} +0i

( i i ) = 0 \Im (i^i) = \boxed{0}

Jun Arro Estrella
Mar 23, 2015

The value i^i is equal to e^(-pi/2) .. thus, taking the imaginary part, which is 0, thus, it is zero..

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