complex number Q (1 Viewer)

bobo123

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(i) if z = x+iy and z'= 1+1/z, obtain an expression for z' in the form x'+iy', and hence express each of x' and y' in terms of x and y.


(ii) find an algebraic relation between x' and y' when z has constant @



dont worry too much about (i) but im really confused about (ii)
i had a go and i got:
k^2*x'=(k+1)*(y')^2-ky'
where k = tan@

am i even doing the right thing here, like by using K?


cheers :)
 
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marsenal

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I'm not sure about the second part but for the first part is it just:

z'= (1+x)/(x<sup>2</sup> + y<sup>2</sup>) - iy/(x<sup>2</sup> + y<sup>2</sup>)

so x'= (1+x)/(x<sup>2</sup> + y<sup>2</sup>)
and y'=-y/(x<sup>2</sup> + y<sup>2</sup>)


Edit: forgot to add the x
 
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wogboy

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i)

z = x + iy

so z' = 1 + 1/z
= 1 + (x - iy)/(x^2 + y^2)
= {1 + x/(x^2 + y^2)} + i {(-y)/(x^2 + y^2)}
or
= 1 + x/|z| - i (y/|z|)

ii)
so

x' = 1 + x/|z|
y' = -y/|z|

given these two equations, we can eliminate |z| to get:

x' = 1 - x/(-y/y')
= 1 - (x/y)y'

and since tan@ = y/x,

x/y = 1/tan@

so,

x' = 1 - y'/tan@
 
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bobo123

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ahh i see i went off into a totally wacked direction

thankyou :)
 

ezzy85

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Originally posted by wogboy
i)

z = x + iy

so z' = 1 + 1/z
= 1 + (x - iy)/(x^2 + y^2)
= {1 + x/(x^2 + y^2)} + i {(-y)/(x^2 + y^2)}
or
= 1 + x/|z| - i (y/|z|)

ii)
so

x' = 1 + x/|z|
y' = -y/|z|

given these two equations, we can eliminate |z| to get:

x' = 1 - x/(-y/y')
= 1 - (x/y)y'

and since tan@ = y/x,

x/y = 1/tan@

so,

x' = 1 - y'/tan@
you put:
z = x + iy

so z' = 1 + 1/z
= 1 + (x - iy)/(x^2 + y^2)

howd you get the (x - iy)/(x^2 + y^2) bit from = 1 + (x - iy)/(x^2 + y^2)
 

Mathematician

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while on the subject...

Someone solve this easy problem( well i think its meant to be)

A Parralelogram ABCD has Diagnols , length 3 and 5. And One side is 2. Find the adjacent side.

I thinkl this is meant to be from the result from complex numbers. The sum of the Square of all the sides of a parralelogram is equal to the sum of the square of this diagnols.

Let adjacent side have length x , the other being 2.

Its a parralelogram.
(opposite sides equal).

2x^2+ 8= 3^2 + 5^2

then find x.


Is this right or am i lost today? lol

And how do u do it the elementary way( without complex numbers , well i havent tried - dont wanna think more on this)
- Wait for ur answers.

:p
 

Mathematician

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oh wait...

:mad1:

The elementary way is the complex no. way ( what was i thinking)

Using cos rule for two triangles ( one triangle has shorter diagnol, the other longer)

gives the complex result.
 

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