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

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Answer 1 · 2016-10-02T06:38:40

Please see below,

As $z = x + i y$ $z=x+iy$

$\frac{i z - 1}{z - i} = \frac{i (x + i y) - 1}{x + i y - i}$ $(iz-1)/(z-i)=(i(x+iy)-1)/(x+iy-i)$

= $\frac{i x - y - 1}{x + i (y - 1)}$ $(ix-y-1)/(x+i(y-1))$

= $\frac{i x - (y + 1)}{x + i (y - 1)} \times \frac{x - i (y - 1)}{x - i (y - 1)}$ $(ix-(y+1))/(x+i(y-1))xx(x-i(y-1))/(x-i(y-1))$

= $\frac{(i x - (y + 1)) (x - i (y - 1))}{x^{2} + {(y - 1)}^{2}}$ $((ix-(y+1))(x-i(y-1)))/(x^2+(y-1)^2)$

= $\frac{i x^{2} + x (y - 1) - x (y + 1) + i (y^{2} - 1)}{x^{2} + {(y - 1)}^{2}}$ $(ix^2+x(y-1)-x(y+1)+i(y^2-1))/(x^2+(y-1)^2)$

= $\frac{x ((y - 1) - (y + 1)) + i (x^{2} + y^{2} - 1)}{x^{2} + {(y - 1)}^{2}}$ $(x((y-1)-(y+1))+i(x^2+y^2-1))/(x^2+(y-1)^2)$

= $\frac{- 2 x + i (x^{2} + y^{2} - 1)}{x^{2} + {(y - 1)}^{2}}$ $(-2x+i(x^2+y^2-1))/(x^2+(y-1)^2)$

As $\frac{i z - 1}{z - i}$ $(iz-1)/(z-i)$ is real

$(x^{2} + y^{2} - 1) = 0$ $(x^2+y^2-1)=0$ and $x^{2} + {(y - 1)}^{2} \neq 0$ $x^2+(y-1)^2!=0$

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

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

Suppose that z = x + yi, where x and y are real numbers. If iz−1z−i(iz-1)/(z-i) is a real number, show that when (x, y) do not equal (0, 1), x2x^2 + y2y^2 = 1?