Question

If `z = x+yi` then what is `(z+i)/(z-i)` in the form `a+bi` ?

Answer 1

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

Explanation:

`(z+i)/(z-i) = (x+(y+1)i)/(x+(y-1)i)`

`color(white)((z+i)/(z-i)) = ((x+(y+1)i)(x-(y-1)i))/((x+(y-1)i)(x-(y-1)i))`

`color(white)((z+i)/(z-i)) = (((x+i)+yi)((x+i)-yi))/(x^2+(y-1)^2)`

`color(white)((z+i)/(z-i)) = ((x+i)^2+y^2)/(x^2+y^2-2y+1)`

`color(white)((z+i)/(z-i)) = (x^2+2x i-1+y^2)/(x^2+y^2-2y+1)`

`color(white)((z+i)/(z-i)) = ((x^2+y^2-1)+2x i)/(x^2+y^2-2y+1)`

`color(white)((z+i)/(z-i)) = (x^2+y^2-1)/(x^2+y^2-2y+1)+(2x)/(x^2+y^2-2y+1) i`

Answer 2

See below.

Explanation:

If `z = x + yi` then `(z+i)/(z-i) = (x+i(y+1))/(x+i(y-1))`

but

`(x+i(y+1))/(x+i(y-1)) = (x+i(y+1))/(x+i(y-1)) xx (x-i(y-1))/(x-i(y-1)) =`

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

Answer 3

`{(x^2+y^2-1)/{x^2+(1-y)^2}}+{(2x)/{x^2+(1-y)^2}}i.`

Explanation:

Given that, `z=x+iy,` we have,

`(z+i)/(z-i)=(x+iy+i)/(x+iy-i),`

`=(x+(1+y)i)/(x-(1-y)i),`

`=(x+(1+y)i)/(x-(1-y)i)xx(x+(1-y)i)/(x+(1-y)i),`

`=[x^2+x{(1+y)i+(1-y)i}+(1+y)i(1-y)i]/[x^2-((1-y)i)^2],`

`={x^2+x(2i)+(1-y^2)i^2}/(x^2-(1-y)^2i^2),`

`={x^2+2ix+(1-y^2)(-1)}/{x^2-(1-y)^2(-1)},`

`=(x^2+y^2-1+2ix)/{x^2+(1-y)^2},`

`={(x^2+y^2-1)/{x^2+(1-y)^2}}+{(2x)/{x^2+(1-y)^2}}i.`

Enjoy Maths!