Eric Nicholas K's answer to lim's Junior College 2 H3 Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Part i, I finally recalled my complex numbers slightly
Date Posted:
4 years ago
Shorter alternative for i) :
(c - a)e^⅓πi
= ce^⅓πi - ae^⅓πi
= re^(θ + 4/3π+⅓π)i - re^(θ + ⅓π)i
= re^(θ+5/3π)i - (- re^(θ + ⅓π + π)i )
= -re^(θ +5/3π - π)i + re^(θ + 4/3π)i
= -re^(θ + ⅔π)i + c
= c - b
(c - a)e^⅓πi
= ce^⅓πi - ae^⅓πi
= re^(θ + 4/3π+⅓π)i - re^(θ + ⅓π)i
= re^(θ+5/3π)i - (- re^(θ + ⅓π + π)i )
= -re^(θ +5/3π - π)i + re^(θ + 4/3π)i
= -re^(θ + ⅔π)i + c
= c - b
360 Tutoring Program