If x,y,b are real,z=x+iy and (z-i)/(z-1)=ib,show that (x-1/2)^2+(y-1/2)^2=1/2?

#(x-1/2)^2+(y-1/2)^2=1/2#

2 Answers
Dec 30, 2017

Please see below.

Explanation:

As #b# is real and#(z-i)/(z-1)=ib#, this means that real part of #(z-i)/(z-1)# is #0#.

Now #(z-i)/(z-1)=(x+iy-i)/(x+iy-1)#

= #(x+i(y-1))/(x-1+iy)# and rationalizing this we get

#((x+i(y-1))(x-1-iy))/((x-1+iy)(x-1-iy))#

= #(x(x-1)+y(y-1)+i(xy-x-y+1-xy))/((x-1)^2+y^2)#

= #(x(x-1)+y(y-1)+i(1-x-y))/((x-1)^2+y^2)#

Hence #(x(x-1)+y(y-1))/((x-1)^2+y^2)=0#

Observe that we cannot have #(1,0)# for #(x,y)# and above means

#x(x-1)+y(y-1)=0#

or #x^2-x+y^2-y=0#

or #x^2-2xx x xx 1/2+(1/2)^2+y^2-2xxyxx1/2+1/4=1/4+1/4#

or #(x-1/2)^2+(y-1/2)^2=1/2#

Jan 1, 2018

Please refer to the Explanation.

Explanation:

Here is another solution to the Problem :

Given that, #"for "x,y,b in RR, and, z=x+iy#,

#(z-i)/(z-1)=ib#.

#:. z-i=ibz-ib, i.e., #

# x+iy-i=ib(x+iy)-ib=ibx+i^2by-ib, or, #

#x+iy-i=ibx-by-ib#.

#:. x+by+i(y-1+b-bx)=0+i0#.

Comparing the Real and Imaginary parts,

# x+by=0, and, y-1+b-bx=0#.

Now, #x+by=0 rArr b=-x/y#.

Sub.ing in the #2^(nd)# eqn., we get,

#y-1+(-x/y)(1-x)=0#.

#:. y^2-y-x+x^2=0, i.e., #

# x^2-x+1/4+y^2-y+1/4=0+1/4+1/4, or, #

# (x-1/2)^2+(y-1/2)^2=1/2#, as desired!

Q.E.D.

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