Willi Yan
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So, i am given this question.
w = -1/2 + √3/2i, w^2 = -1/2 - √3/2i
First,I was asked to find w and w^2 in modulus argument form which are (1, 2π/3) and (1, -2π/3) respectively.
Next, I was asked to draw an Argand diagram to explain why 1 + w + w^2 = 0. This is where got stuck.
What I've drawn so far is a +1 on positive x-axis and then, w continuing above x axis at an angle of 2π/3 from x-axis. After that comes w^2 continuing from the last point to the origin which finally results in a 0 in terms of displacement. Literally, it forms an equilateral triangle. However, in terms of magnitude, it results in 1 (1 from x axis + 1 from w - 1 from w^2 gives a resultant of 1) while it should be 0. I am willing to listen to any explanation which explains how to draw this diagram completly.
Thank you.
Last edited by Willi Yan; 1 year ago
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mqb2766
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(Original post by Willi Yan)
So, i am given this question.
w = -1/2 + √3/2i, w^2 = -1/2 - √3/2i
First,I was asked to find w and w^2 in modulus argument form which are (1, 2π/3) and (1, -2π/3) respectively.
Next, I was asked to draw an Argand diagram to explain why 1 + w + w^2 = 0. This is where got stuck.
What I've drawn so far is a +1 on positive x-axis and then, w continuing above x axis at an angle of 2π/3 from x-axis. After that comes w^2 continuing from the last point to the origin which finally results in a 0 in terms of displacement. Literally, it forms an equilateral triangle. However, in terms of magnitude, it results in 1 (1 from x axis + 1 from w - 1 from w^2 gives a resultant of 1) while it should be 0. I am willing to listen to any explanation which explains how to draw this diagram completly.
Thank you.
w and w^2 are conjugate so their sum is twice their real values.
You should recognise they're cube roots of unity?
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Willi Yan
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(Original post by mqb2766)
w and w^2 are conjugate so their sum is twice their real values.
You should recognise they're cube roots of unity?
I'm sorry i don't quite get the first statement. Is it that their sum which is (1+1 = 2) is twice their real values which is 1 for both w and w^2.
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mqb2766
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(Original post by Willi Yan)
I'm sorry i don't quite get the first statement. Is it that their sum which is (1+1 = 2) is twice their real values which is 1 for both w and w^2.
No, their real parts are -1/2. Just look at the rectangular form.
You can also get the relationship from basic algebra from
w^3 = 1.
There are a few ways to do it.
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Willi Yan
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I'm sorry it just looks messy but here's what i drew and another is what's shown in the mark scheme. I think my drawing is correct.
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mqb2766
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(Original post by Willi Yan)
I'm sorry it just looks messy but here's what i drew and another is what's shown in the mark scheme. I think my drawing is correct.
Yes
w+w^2 = -1
So
-1+1=0
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Willi Yan
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(Original post by mqb2766)
Yes
w+w^2 = -1
So
-1+1=0
alright thank you very much!!!
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