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Unread 19 Apr 2004, 13:52   #1
'|'empest
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[Maths] Complex number question

I havn't been able to find a single example like it on google, which worries me a tad.

find all z € C which satisfy the equation

7z - 3z' = 1


nb z' should be z bar, i.e the conjugate


i've got as far as 4a + 10bi + i^2 = 0

i've gone horribly wrong havn't i?
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Unread 19 Apr 2004, 14:02   #2
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Re: [Maths] Complex number question

Let z=x+iy, z'=x-iy.
7z - 3z' = 4x + 10iy = 1
x=1/4, y=0
z=1/4
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Unread 19 Apr 2004, 14:04   #3
'|'empest
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Re: [Maths] Complex number question

is it really that simple?

i have myself sometimes...

thanks yo
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Unread 19 Apr 2004, 14:04   #4
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Re: [Maths] Complex number question

7(a+ib) - 3(a-ib) = 1 fair enough?
7a + 7ib -3a +3ib = 1 fair enough?
4a + 10ib = 1 k?

then 4a = 1 and 10ib = 0,
so a = 0.25 and b = 0, z = 0.25

i dunno what you are doing. maybe i missed something.
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Unread 19 Apr 2004, 14:07   #5
'|'empest
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Re: [Maths] Complex number question

i brought 1 over to the left, -1 = i^2

i was looking for a much harder answer than they're after apparently...
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Unread 19 Apr 2004, 14:13   #6
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Re: [Maths] Complex number question

Quote:
Originally Posted by '|'empest
i brought 1 over to the left, -1 = i^2

i was looking for a much harder answer than they're after apparently...
i^2 + 10b i + 4a = 0
i = ( -10b ± sqrt(100b^2 - 16a) ) / 2
i = -5b ± sqrt(25b^2 - 4a)
equating real parts: b = 0
square both sides: -4a = -1
a = 1/4
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Unread 19 Apr 2004, 14:29   #7
'|'empest
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Re: [Maths] Complex number question

bugger
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