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?
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.
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.
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 ??
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?