# We have to show that the function S: C\{1} -> C\{1} with S(z) = (z+i)/(z+1) is a bijection (C=complex numbers)How would we decompose z in this situation?

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You need to write the complex number such that: `z = a+bi` .

Hence, `z + i = a + bi+ i =gt z + i = a+i(b+1)` and `z+1 = (a+1)+bi`

Writing the new form of the function yields:

`S(z) = (a+i(b+1))/((a+1)+bi)`

You need to multiply by the conjugate of `((a+1)+bi)` such that:

`S(z) =((a+i(b+1))*((a+1)-bi))/((a+1)^2+b^2)`

`S(z) = (a(a+1) - abi + i*(a+1)(b+1) + b(b+1))/((a+1)^2+b^2)`

`S(z) = (a(a+1) + i*(a+b+1) + b(b+1))/((a+1)^2+b^2)`

Hence, you need to prove that S(z) is bijection. You know the rule of the function: `S(z) = (a(a+1)+b(b+1)+i(a+b+1))/((a+1)^2+b^2)`

The function is a bijection if it is both injective and surjective.

1. You need to prove that the function is injective, hence if `f(x)=f(y)=gtx=y` , such that:

`S(z)=S(a+bi)=S(c+di)`

`(a(a+1)+b(b+1)+i(a+b+1))/((a+1)^2+b^2)=(c(c+1)+d(d+1)+i(c+d+1))/((c+1)^2+d^2)`

Isolating the real parts and the imaginary parts both sides yields:

`(a(a+1)+b(b+1))/((a+1)^2+b^2) + (i(a+b+1))/((a+1)^2+b^2) = (c(c+1)+d(d+1))/((c+1)^2+d^2) + (i(c+d+1))/((c+1)^2+d^2)`

`(a(a+1)+b(b+1))/((a+1)^2+b^2) - (c(c+1)+d(d+1))/((c+1)^2+d^2) = i*((c+d+1)/((c+1)^2+d^2) - (a+b+1)/((a+1)^2+b^2))`

`i = [(a(a+1)+b(b+1))/((a+1)^2+b^2) - (c(c+1)+d(d+1))]/((c+1)^2+d^2)/((c+d+1)/((c+1)^2+d^2) - (a+b+1)/((a+1)^2+b^2))`

Notice that yields a contradiction since `i = sqrt(-1), ` hence`S(a+bi)=S(c+di),` hence S(z) is injective.

2. You need to prove that S(z) is surjective, hence you need to prove that for any complex number `z_2` in the range `C-{1}` there is `z_1` , a complex number in domain `C-{-1}, ` such that `f(z_2)=z_1`

Hence, for any comlplex number `z_1 = a+bi` , there is a conjugate `z_2 = a - bi` , such that `f(z_2) = z_1` .

**Hence, the steps 1. and 2. proves that function S(z) is a bijection over complex numbers set C.**

Sorry , type mistake: the definition range must read:

S: C\{-1} -> c\{+1}