Are real roots considered complex roots (James Ruse 2013 Task 1 solutions) (1 Viewer)

[trea99]

Hey guys,
I've been doing a few past MX2 papers on complex numbers and I often see in the worked solutions that real roots such 1 are not considered complex.
Like for cube root of unity a solution I saw for the question "if w is a complex cube root of unity, prove that w^2 is also a root" was this,
"if w is a complex cube root of unity,
w^3 =1
(w^2)^3 = (w^3)^2
= 1^2
= 1
Assuming w^2 = w, then w(w-1) = 0 and w = 0 or 1
The bit that confuses me is here they say,
But w is given to be a complex root so w does not = 0 or 1 and thus w^2 is a different non real root.
Isn't this line saying that non real and complex are the same. I though that complex numbers included real ones but only imaginary numbers were non real. Can someone please explain. Is this a valid solution (its provided in James Ruse 2013 Task 1 btw)
Thank you!

 

[pikachu975]

Hey guys,
I've been doing a few past MX2 papers on complex numbers and I often see in the worked solutions that real roots such 1 are not considered complex.
Like for cube root of unity a solution I saw for the question "if w is a complex cube root of unity, prove that w^2 is also a root" was this,
"if w is a complex cube root of unity,
w^3 =1
(w^2)^3 = (w^3)^2
= 1^2
= 1
Assuming w^2 = w, then w(w-1) = 0 and w = 0 or 1
The bit that confuses me is here they say,
But w is given to be a complex root so w does not = 0 or 1 and thus w^2 is a different non real root.
Isn't this line saying that non real and complex are the same. I though that complex numbers included real ones but only imaginary numbers were non real. Can someone please explain. Is this a valid solution (its provided in James Ruse 2013 Task 1 btw)
Thank you!

Complex numbers are the biggest field of numbers and real numbers and imaginary are included in it. Check out the history of maths numbers to understand.

Real means there is no imaginary part
Non real means there is no real part
Imaginary means there is an imaginary part

 

[trea99]

Yep, I understand that but the question states the w is a complex cube root of unity which (in my opinion at least) does not imply that
w =1 is not a solution. However, that is the line of evidence that the provided solution uses as to why w^2 is also a cube root of unity. The solutions consider w to take only 2 values, that is the roots of 1 + w + w^2 = 0, cis2pi/3 and cis4pi/3 where if either is w then the other is w^2. This is not true if w = 1 which is what I'm not understanding. I've seen other questions word it as w is an imaginary cube root of unity or w is the complex cube root of unity with the smallest positive argument both of which make more sense as to why w can not equal 1. But if its merely given that w is complex, why is it that w can not equal 1 in this case? Or is the question just worded wrong and it should be treated as if w were purely imaginary?
Question:
3c) "If w is one of the complex cube roots of unity, prove that the other complex root is w^2"
Solution given:
w^3 = 1 (given w is a cube root of unity)
(w^2)^3 = (w^3)^2
= 1^2
= 1
Therefore, w^2 is also a root
Assume that w^2 = w
Then w(w-1) = 0
w = 0 or 1
Not possible since w is non real
Hence w^2 is a different non real root.
My question is have they just mixed up complex and non real and considered both to be identical when writing this question or am I missing something?

 

[si2136]

Yep, I understand that but the question states the w is a complex cube root of unity which (in my opinion at least) does not imply that
w =1 is not a solution. However, that is the line of evidence that the provided solution uses as to why w^2 is also a cube root of unity. The solutions consider w to take only 2 values, that is the roots of 1 + w + w^2 = 0, cis2pi/3 and cis4pi/3 where if either is w then the other is w^2. This is not true if w = 1 which is what I'm not understanding. I've seen other questions word it as w is an imaginary cube root of unity or w is the complex cube root of unity with the smallest positive argument both of which make more sense as to why w can not equal 1. But if its merely given that w is complex, why is it that w can not equal 1 in this case? Or is the question just worded wrong and it should be treated as if w were purely imaginary?
Question:
3c) "If w is one of the complex cube roots of unity, prove that the other complex root is w^2"
Solution given:
w^3 = 1 (given w is a cube root of unity)
(w^2)^3 = (w^3)^2
= 1^2
= 1
Therefore, w^2 is also a root
Assume that w^2 = w
Then w(w-1) = 0
w = 0 or 1
Not possible since w is non real
Hence w^2 is a different non real root.
My question is have they just mixed up complex and non real and considered both to be identical when writing this question or am I missing something?

Any real number is a complex number. For example, 6 = 6 + 0i. Technically, there is still an imaginary component to it. Hence, you can complete the square with complex numbers.

For example: (z^2 + 2z + 2) = 0

z^2 + 2z +1 +1 = 0

(z+1)^2 - i^2 = 0 (Notice REAL 1 into IMAGINARY -I^2)

Therefore, (z + 1 - i)(z +1 + i)=0

 

[InteGrand]

Yep, I understand that but the question states the w is a complex cube root of unity which (in my opinion at least) does not imply that
w =1 is not a solution. However, that is the line of evidence that the provided solution uses as to why w^2 is also a cube root of unity. The solutions consider w to take only 2 values, that is the roots of 1 + w + w^2 = 0, cis2pi/3 and cis4pi/3 where if either is w then the other is w^2. This is not true if w = 1 which is what I'm not understanding. I've seen other questions word it as w is an imaginary cube root of unity or w is the complex cube root of unity with the smallest positive argument both of which make more sense as to why w can not equal 1. But if its merely given that w is complex, why is it that w can not equal 1 in this case? Or is the question just worded wrong and it should be treated as if w were purely imaginary?
Question:
3c) "If w is one of the complex cube roots of unity, prove that the other complex root is w^2"
Solution given:
w^3 = 1 (given w is a cube root of unity)
(w^2)^3 = (w^3)^2
= 1^2
= 1
Therefore, w^2 is also a root
Assume that w^2 = w
Then w(w-1) = 0
w = 0 or 1
Not possible since w is non real
Hence w^2 is a different non real root.
My question is have they just mixed up complex and non real and considered both to be identical when writing this question or am I missing something?

Not sure why the solutions keep saying stuff once they've shown that (w^2)^3 is equal to 1. Once you show this, you are done. (In fact similar reasoning shows that if w is any n-th root of unity, then so is w^k for any integer k). Maybe they meant to ask something else than what they actually wrote.