Question

find the bilinear transformation which maps the points 0,-i,-1 to i,1,0​

Answer 1

Step-by-step explanation:

Bilinear transformation which maps the point (1,i,−1) onto (0,1,∞) is 8 σ (a) ω=

z+1

iz−i

(b)ω=

r+1

L

(c)ω=

z+1

−iz+i

(d) None of the above

Answer 2

Answer:

The bilinear transformation which maps the points $0,-i,-1$ to $i,1,0$ is $w=\frac{z+1}{i(z-1)}$.

Step-by-step explanation:

Given: The points $z_1=0$, $z_2=-i$ and $z_3=-1$ are mapped to $w_1=i$, $w_2=1$ and $w_3=0$ respectively.

To find: The bilinear transformation between the given points.

Step 1 of 2

Let the transformation be,

$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ . . . . . (1)

Substitute all the values in the expression (1) as follows:

⇒ $\frac{(w-i)(1-0)}{(w-0)(1-i)}=\frac{(z-0)(-i-(-1))}{(z-(-1))(-i-0)}$

Simplify as follows:

⇒ $\frac{(w-i)(1-0)}{(w-0)(1-i)}=\frac{(z-0)(-i+1)}{(z+1)(-i-0)}$

⇒ $\frac{(w-i)(1)}{w(1-i)}=\frac{(z)(-i+1)}{(z+1)(-i)}$

Step 2 of 2

Find the value of $w$.

⇒ $\frac{(w-i)}{w(1-i)}=\frac{z(-i+1)}{-i(z+1)}$

Cross multiply as follows:

⇒ $-i(w-i)(z+1)=zw(-i+1)(1-i)$

⇒ $-i(w-i)(z+1)=zw(-i+1)^2$

Compute the products as follows:

⇒ $-i(wz+w-iz-i)=zw(-1+1-2i)$

⇒ $-iwz-iw+i^2z+i^2=-zw+zw-2zwi$

⇒ $-iwz-iw-z-1=-2zwi$ (Since $i^2=-1$)

Further, simplify as follows:

⇒ $-iwz+2zwi-iw=z+1$

⇒ $zwi-iw=z+1$

⇒ $w(zi-i)=z+1$

⇒ $w=\frac{z+1}{zi-i}$

or,

⇒ $w=\frac{z+1}{i(z-1)}$

Final answer: The bilinear transformation which maps the points $0,-i,-1$ to $i,1,0$ is $w=\frac{z+1}{i(z-1)}$.

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