Suppose that z = x + yi, where x and y are real numbers. If $(iz-1)/(z-i)$ is a real number, show that when (x, y) do not equal (0, 1), $x^2$ + $y^2$ = 1?

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Answer 1 · 2016-10-02T06:38:40

Please see below,

As $z=x+iy$

$(iz-1)/(z-i)=(i(x+iy)-1)/(x+iy-i)$

= $(ix-y-1)/(x+i(y-1))$

= $(ix-(y+1))/(x+i(y-1))xx(x-i(y-1))/(x-i(y-1))$

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

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

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

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

As $(iz-1)/(z-i)$ is real

$(x^2+y^2-1)=0$ and $x^2+(y-1)^2!=0$

Now as $x^2+(y-1)^2$ is sum of two squares, it can be zero only when $x=0$ and $y=1$ i.e.

if $(x,y)$ is not $(0,1)$ , $x^2+y^2=1$

Suppose that z = x + yi, where x and y are real numbers. If (iz-1)/(z-i)(iz-1)/(z-i) is a real number, show that when (x, y) do not equal (0, 1), x^2x^2 + y^2y^2 = 1?