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#