Bilinear transformation which maps the point \[(1,i, - 1)\] onto \[(0,1,\infty )\] is
a) \[w = \dfrac{{iz - i}}{{z + 1}}\]
b) \[w = \dfrac{{iz}}{{z + 1}}\]
c) \[w = \dfrac{{ - iz + i}}{{z + 1}}\]
d) None of the above

Hint: Here the question is related to the bilinear transformation. We have to find the function of transformation where the points of z and the value of w are given. By using the formula \[w = \dfrac{{az + b}}{{cz + d}}\], here the point \[(1,i, - 1)\] represents the value of z and \[(0,1,\infty )\] represents the value of w. Using the formula and on further simplification we have to determine the value of a, b ,c and d. Then we have to substitute these values in the formula. Hence, we obtain the required solution.

Complete step by step solution:
Bilinear transform , a type of conformal map used to switch between continuous-time and discrete-time representations.
The formula for the bilinear transformation is given by
 \[w = \dfrac{{az + b}}{{cz + d}}\] ----- (1)
Now consider the given question
The point \[(1,i, - 1)\] will represents the value of z and the point \[(0,1,\infty )\] will represents the value of w for the value of z respectively.,
When \[z = 1\], then the value of w will be \[w = 0\]
On substituting these values in the equation (1) we have
\[ \Rightarrow 0 = \dfrac{{a(1) + b}}{{c(1) + d}}\]
On multiplying the terms
\[ \Rightarrow 0 = \dfrac{{a + b}}{{c + d}}\]
Take (c + d) to the LHS we get
\[ \Rightarrow 0\, \times \,(c + d) = a + b\]
The product of any number and 0 will be zero.
 \[ \Rightarrow 0 = a + b\]
Take b to LHS we have
\[ \Rightarrow a = - b\]----- (2)
When \[z = i\], then the value of w will be \[w = 1\]
On substituting these values in the equation (1) we have
\[ \Rightarrow 1 = \dfrac{{a(i) + b}}{{c(i) + d}}\]
On multiplying the terms
\[ \Rightarrow 1 = \dfrac{{ai + b}}{{ci + d}}\]
Take \[(ci + d)\] to the LHS we get
\[ \Rightarrow 1\, \times \,(ci + d) = ai + b\]
On multiplying we have
\[ \Rightarrow ci + d = ai + b\]
Take \[ci\] to the RHS and \[b\] to LHS, we have
\[ \Rightarrow d - b = ai - ci\]------- (3)
When \[z = - 1\], then the value of w will be \[w = \infty \]
On substituting these values in the equation (1) we have
\[ \Rightarrow \infty = \dfrac{{a( - 1) + b}}{{c( - 1) + d}}\]
On multiplying the terms
\[ \Rightarrow \infty = \dfrac{{ - a + b}}{{ - c + d}}\]
Take \[\infty \] to the RHS and \[( - c + d)\] to RHS we get
\[ \Rightarrow ( - c + d) = \dfrac{{ - a + b}}{\infty }\]
When the number is divided by infinity then the answer will be zero.
 \[ \Rightarrow - c + d = 0\]
Take c to RHS we have
\[ \Rightarrow d = c\]----- (4)
Now we have obtained 3 equations.
\[ \Rightarrow a = - b\]----- (2)
\[ \Rightarrow d - b = ai - ci\]------- (3)
\[ \Rightarrow d = c\]----- (4)
On substituting the equation (2) and equation (4) in the equation (3) we have
\[ \Rightarrow c + a = ai - ci\]
Take a to the RHS and ci to the LHS we have
\[ \Rightarrow ci + c = ai - a\]
Take a and c as common
\[ \Rightarrow c(i + 1) = a(i - 1)\]
This can be written as
\[ \Rightarrow \dfrac{a}{c} = \dfrac{{i + 1}}{{i - 1}}\]
To the RHS we multiply both numerator and denominator by the conjugate of \[i - 1\]. The conjugate of \[i - 1\] is \[i + 1\]
On multiplying we have
\[ \Rightarrow \dfrac{a}{c} = \dfrac{{i + 1}}{{i - 1}} \times \dfrac{{i + 1}}{{i + 1}}\]
\[ \Rightarrow \dfrac{a}{c} = \dfrac{{{{\left( {i + 1} \right)}^2}}}{{{i^2} - 1}}\]
\[ \Rightarrow \dfrac{a}{c} = \dfrac{{{i^2} + 1 + 2i}}{{{i^2} - 1}}\]
As we know that the value of \[{i^2} = - 1\]
\[ \Rightarrow \dfrac{a}{c} = \dfrac{{ - 1 + 1 + 2i}}{{ - 1 - 1}}\]
On simplifying we have
\[ \Rightarrow \dfrac{a}{c} = \dfrac{{2i}}{{ - 2}}\]
On cancelling 2 we have
\[ \Rightarrow \dfrac{a}{c} = - i\]
\[ \Rightarrow a = - ic\]---- (5)
Now we will consider the bilinear transformation formula
\[ \Rightarrow w = \dfrac{{az + b}}{{cz + d}}\]
As we know that \[b = - a\] and \[c = d\], now the formula will be written as
\[ \Rightarrow w = \dfrac{{az - a}}{{cz + c}}\]
By the equation (5) the above term is written as
\[ \Rightarrow w = \dfrac{{( - ic)z - ( - ic)}}{{cz + c}}\]
on simplifying we have
\[ \Rightarrow w = \dfrac{{ - icz + ic}}{{cz + c}}\]
Take c as a common from the both numerator and denominator we have
\[ \Rightarrow w = \dfrac{{c( - iz + i)}}{{c(z + 1)}}\]
On cancelling we have
\[ \Rightarrow w = \dfrac{{ - iz + i}}{{z + 1}}\]
Hence the option c is the correct one.
So, the correct answer is “Option C”.

Note: In this question we have to determine the four unknowns that are a, b, c and d. To find these we should have 4 equations. In general if we are determining we should have n equations. Here especially in this question the equation (2), (3) (4) and (5) are the four equations used to determine the four unknowns. Otherwise we should determine the value of unknowns in the terms of another unknown and it should be linked to each unknown term.