Find the image of |z-2i|=2 under transformation w=1/z.

Описание к видео Find the image of |z-2i|=2 under transformation w=1/z.

The transformation w=1/z transforms all circles and straight lines in the z-plane into circles or straight lines in the w-plane.

w=1/z
i.e., z=1/w
then z=1/(u+iv)
=1/(u+iv) × (u-iv)/(u-iv)
=u-iv/(u^2+v^2)

i.e., x+iy = u/(u^2+v^2) + i[-v/(u^2+v^2)]

then we will get, x= u/(u^2+v^2) &
y= -v/(u^2+v^2)

We have the general equation
a(x^2+y^2)+2gx+2fy+c=0

then
a [ u^2/(u^2+v^2)^2 + v^2/(u^2+v^2)^2 ] +2g[ u/(u^2+v^2) ] -2f[ v/(u^2+v^2) ] +c =0

a (u^2+v^2)/(u^2+v^2)^2 +2g u/(u^2+v^2) -2f v/(u^2+v^2) + c=0

We will get the transfromed equation,
c(u^2+v^2) +2gu -2fv+a=0

1) If a≠0, c≠0
then
circles not passing through the origin in Z-plane map into circles not passing through the origin in the W-plane.

2) If a≠0, c=0
then
circles through the origin in Z-plane map onto straight lines not through the origin in the W-plane.

3) If a=0, c≠0
then
the straight lines not through the origin in Z-plane map onto circles through the origin in the W-plane.

4) If a=0,c=0
then
straight lines throught thr origin of Z-plane onto straight lines through the origin in the W-plane.




A complex number Z is an ordered pair (x,y) of real numbers x,y and we write z=(x,y).

and write Re z = x,
Im z = y.

We call x ' the real part of z and y the imaginary part of z'.

*In our problem we have centre z0=2i=0+i2=(0,2).



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