mathsbrain
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How do you graph |z-1/z|=2?
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help plz (3 Viewers)
- Thread starter mathsbrain
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Eagle Mum
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I’m not a maths expert, but nobody else has responded, so this is how I would approach this question:
As a maths extn 2 question, I gather that ‘z’ is a complex number of the form x + iy
|z - 1/z| = 2
z - 1/z = 2 or z - 1/z = -2
which produces two quadratic equations:
z^2 - 2z - 1 = 0 and z^2 + 2z - 1 = 0
and four possible values of z and therefore x (since z has no imaginary component in any of these solutions):
1 + sqrt (2) and -1 + sqrt (2)
Hence, I would plot these values on the x - axis with y = 0 for each point.
I don’t know how I would prove there are no solutions with non zero values of y, nor how I would find such solutions if they exist.
As a maths extn 2 question, I gather that ‘z’ is a complex number of the form x + iy
|z - 1/z| = 2
z - 1/z = 2 or z - 1/z = -2
which produces two quadratic equations:
z^2 - 2z - 1 = 0 and z^2 + 2z - 1 = 0
and four possible values of z and therefore x (since z has no imaginary component in any of these solutions):
1 + sqrt (2) and -1 + sqrt (2)
Hence, I would plot these values on the x - axis with y = 0 for each point.
I don’t know how I would prove there are no solutions with non zero values of y, nor how I would find such solutions if they exist.
This is a really tough question and I'm sure there's some niche way to solve it, but brute forcing it by subbing z=x+iy is always an option
^2+4x^2y^2}=2\sqrt{x^2+y^2})
Now I found this factorisation on the internet, but you definitely wouldn't be able to find this normally
^2+y^2-2\right) \left((x+1)^2+y^2-2\right)=0)
By null factor theorem, you get 2 equations:
^2+y^2\right)=2)
^2+y^2\right)=2)
The graph is 2 circles, centred at (0, 1) and (0, -1), both with radius root 2
Now I found this factorisation on the internet, but you definitely wouldn't be able to find this normally
By null factor theorem, you get 2 equations:
The graph is 2 circles, centred at (0, 1) and (0, -1), both with radius root 2