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How do you manually calculate i to the power of i? - Answers

i (taken to be sqrt(-1) for this question) requires that you know a bit about writing complex numbers. i = e^(i*pi/2) so i^i = (e^(i*pi/2))^i which equals e^(i*i*pi/2) since i*i = -1 we get e^(-pi/2) so i^i = e^(-pi/2) which is roughly .207879576



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How do you manually calculate i to the power of i? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_manually_calculate_i_to_the_power_of_i

i (taken to be sqrt(-1) for this question) requires that you know a bit about writing complex numbers. i = e^(i*pi/2) so i^i = (e^(i*pi/2))^i which equals e^(i*i*pi/2) since i*i = -1 we get e^(-pi/2) so i^i = e^(-pi/2) which is roughly .207879576



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https://math.answers.com/math-and-arithmetic/How_do_you_manually_calculate_i_to_the_power_of_i

How do you manually calculate i to the power of i? - Answers

i (taken to be sqrt(-1) for this question) requires that you know a bit about writing complex numbers. i = e^(i*pi/2) so i^i = (e^(i*pi/2))^i which equals e^(i*i*pi/2) since i*i = -1 we get e^(-pi/2) so i^i = e^(-pi/2) which is roughly .207879576

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      i (taken to be sqrt(-1) for this question) requires that you know a bit about writing complex numbers. i = e^(i*pi/2) so i^i = (e^(i*pi/2))^i which equals e^(i*i*pi/2) since i*i = -1 we get e^(-pi/2) so i^i = e^(-pi/2) which is roughly .207879576

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