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

3 Answers
Oct 19, 2017

#(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#

Oct 19, 2017

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)#

Oct 19, 2017

# {(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!