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junior college 2 | H3 Maths
6 Answers Below
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Help me please.deeply appreciate.
Date Posted: 4 years ago
Views: 397
Hang on, attempting part i
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I am only capable of doing this one part. Forgot my complex numbers fully.
Date Posted:
4 years ago
thanks for yout hard effort，really appreciate
Shorter alternative:
a² + b² + c² = bc + ca + ab
(-4 + 4i)² + (4 - 2i)² + c² = (4 - 2i)c + c(-4 + 4i) + (-4 + 4i)(4 - 2i)
16 - 32i - 16 + 16 - 16i - 4 + c² = c(4 - 2i - 4 + 4i) - 16 + 8i + 16i + 8
12 - 48i + c² = 2ci - 8 + 24i
c² - 2ci - 1 = 72i - 21
c² - 2ci + i² = 3(9 - 24i - 16)
(c - i)² = 3(3² - 2(3)(4i) + (4i)²)
(c - i)² = 3(3 + 4i)²
k = 3
a² + b² + c² = bc + ca + ab
(-4 + 4i)² + (4 - 2i)² + c² = (4 - 2i)c + c(-4 + 4i) + (-4 + 4i)(4 - 2i)
16 - 32i - 16 + 16 - 16i - 4 + c² = c(4 - 2i - 4 + 4i) - 16 + 8i + 16i + 8
12 - 48i + c² = 2ci - 8 + 24i
c² - 2ci - 1 = 72i - 21
c² - 2ci + i² = 3(9 - 24i - 16)
(c - i)² = 3(3² - 2(3)(4i) + (4i)²)
(c - i)² = 3(3 + 4i)²
k = 3
thanks for your alternative method
Welcome. Do take note of the need to reject the value of c with both real and imaginary part negative in the last part
deeply appreciate
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Last part.
For the first three parts, perhaps J or Boy Mow Chau may have a better thought on this. I have a rough idea for the first two parts geometrically, but have no idea on how to put them in complex form.
For the first three parts, perhaps J or Boy Mow Chau may have a better thought on this. I have a rough idea for the first two parts geometrically, but have no idea on how to put them in complex form.
Date Posted:
4 years ago
The negative result has to be rejected as the question states that the equilateral triangle is labelled anticlockwise.
This negative result would make the triangle become labelled clockwise instead.
This negative result would make the triangle become labelled clockwise instead.
(c - i)² = 3(3 + 4i)² = (√3)²(3 + 4i)²
(c - i)² = (3√3 + 4√3 i)²
(c - i)² = (3√3 + (4√3 + 1)i - i)²
Comparing coefficients,
c = 3√3 + (4√3 + 1)i
(c - i)² = (3√3 + 4√3 i)²
(c - i)² = (3√3 + (4√3 + 1)i - i)²
Comparing coefficients,
c = 3√3 + (4√3 + 1)i
Interesting...
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Alternative for last part, using the usual long-winded method.
Date Posted:
4 years ago
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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
done
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Part ii
Date Posted:
4 years ago
(c - a)e^-⅓πi
= ce^-⅓πi - ae^-⅓πi
= re^(θ + 4/3π - ⅓π)i - re^(θ - ⅓π)i
= re^(θ+π)i - (- re^(θ - ⅓π + π)i )
= -re^(θ + π - π)i + re^(θ + ⅔π)i
= -re^θi + b
= b - a
= ce^-⅓πi - ae^-⅓πi
= re^(θ + 4/3π - ⅓π)i - re^(θ - ⅓π)i
= re^(θ+π)i - (- re^(θ - ⅓π + π)i )
= -re^(θ + π - π)i + re^(θ + ⅔π)i
= -re^θi + b
= b - a
done
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Third part
Date Posted:
4 years ago
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