Question

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`

Answer 1

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`

Answer 2

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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