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Tranformations from z to w plane

Need a bit of help with this please.

From z = x + iy to w = u + iv, transformation T given by w = 1/z. Show that the image, under T, of the line with equation x = 1/2 in the Z plane is a circle C in the w plane. Find equation of C.

I can't get to first base here. I can do the ones where |z| = 2, or if you have a circle in the Z plane, but I cannot formulate the x = 1/2 - the only thing I can try to do is think about making up the perpendicular bisector to simuklate the locus, such that |z| = |z +1|, but not sure if that helps or not?

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Reply 1
letting x = 1/2

you get w = 1/( 1/2 + iy ) now multiply top & bottom by the conjugate of the bottom...

rearrange to get

2/(4y2 + 1) - 4iy/(4y2 + 1) **

now choose some sensible values for y and plot them... you should be able to find the centre and radius of the circle. Then go back and show that the u + vi from ** fit the formula for the circle you have found.
Reply 2
Thanks - a little tough - I was hoping I might be able to show something that actually looked like a circle! (at least to me)
Reply 3
I hope this helps....
Reply 4
Original post by Charries
Thanks - a little tough - I was hoping I might be able to show something that actually looked like a circle! (at least to me)


w=1z    z=1w \displaystyle w = \frac1{z} \implies z = \frac1{w}

Sub in w=u+iv w = u + iv , them multiply top and bottom by the conjugate.

You will get, z=uivu2+v2=uu2+v2ivu2+v2 \displaystyle z = \frac{u-iv}{u^2 +v^2} = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

We know z=x+iy=12+iy z = x + iy = \dfrac12 + iy

12+iy=uu2+v2ivu2+v2 \displaystyle \frac12 + iy = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

Equate the real parts, and you should get the equation of circle.

Sorry for showing too much working.
Reply 5
Original post by mikelbird
I hope this helps....


Isn't my way better?
Reply 6
I would say its the same...eliminating between the variables is the only really complicated part and both of us would have to do it..
Reply 7
Original post by mikelbird
I would say its the same...eliminating between the variables is the only really complicated part and both of us would have to do it..


Next steps in my method are,

12+iy=uu2+v2ivu2+v2 \displaystyle \frac12 + iy = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

Equate real parts,
12=uu2+v2    (u1)2+v2=1 \displaystyle \frac12 = \frac{u}{u^2 + v^2} \implies (u-1)^2 + v^2 = 1

I think that your method is quite complex.
Reply 8
Original post by raheem94
Isn't my way better?


Actually , on second thoughts...it can very much depend on the question....sometimes its easier your way...sometimes mine...!!
Reply 9
Thanks to you both. Helps a lot.
Reply 10
... follow up question.

What if y = 2x + 1 instead - how woud that mapping work?
Reply 11
Original post by Charries
... follow up question.

What if y = 2x + 1 instead - how woud that mapping work?


x+iy=uu2+v2ivu2+v2 \displaystyle x + iy = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

The above is taken from one of my previous post in this thread.

Equate the real and imaginary parts,

x=uu2+v2  and  y=vu2+v2 \displaystyle x = \frac{u}{u^2 + v^2} \ \ and \ \ y = - \frac{v}{u^2+v^2}

Insert it in the equation y=2x+1 y = 2x +1

vu2+v2=2×uu2+v2+1 \displaystyle - \frac{v}{u^2+v^2} = 2 \times \frac{u}{u^2 + v^2} + 1

Now simplify it and you will get an equation of a circle.
Reply 12
Raheen94 would give you the answer...
Reply 13
Thank you guys - got it.
Reply 14
Ok - I have done a few questions like this and everything is going well, but come across this one. w = 16/z, where |z-4|=4.

Gone through the usual method, converted the thing to a circle in cartesian co-ordinates and get some awful mess of a squared number once I compare real and imaginary parts and substiture for x and y into (what I get) as (u-4)^2 + v^2 = 16.

Any further clues?
Reply 15
z=16wz=\frac{16}{w} and z4=4|z-4|=4

16w4=4\left|\frac{16}{w}-4\right|=4

4w1=1\left|\frac{4}{w}-1\right|=1

4ww=1\left|\frac{4-w}{w}\right|=1

4w=w|4-w|=|w|

w=2+viw=2+vi
Reply 16
Original post by BabyMaths
z=16wz=\frac{16}{w} and z4=4|z-4|=4

16w4=4\left|\frac{16}{w}-4\right|=4

4w1=1\left|\frac{4}{w}-1\right|=1

4ww=1\left|\frac{4-w}{w}\right|=1

4w=w|4-w|=|w|

w=2+viw=2+vi


Nice one...and there is a geometric aspect to this...that last equation before the end is telling you that the locus of the point z is such that it is always equidistant from the point 0 and the point 4 i.e. it is the perpendicular bisector of the two points.

Anyone like to guess (without calculation) what locus is |z| +|z-4| = 6 ??
Reply 17
Original post by mikelbird


Anyone like to guess (without calculation) what locus is |z| +|z-4| = 6 ??


Thanks. :smile:

Anyone who knows the construction using two pins and a piece of string will know what it is.
Reply 18
That's brilliant - thank you.

photo(20).JPG

In this mark scheme, have they just got rid of the i because it's a factor of both, and they can pull it out into another modulus and say it's 1. I can see why mod of i squared would be 1 and can be ignored, but I am not convinced why they can do it for i?
Reply 19
Actually please explain
(edited 11 years ago)

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