Question

the invariant points of the transformation W=(3z-5i)/(iz-1)​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Transformation is}$

$\mathsf{W=\dfrac{3z-5i}{iz-1}}$

$\underline{\textbf{To find:}}$

$\textsf{Invariant points of the given transformation}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{W=\dfrac{3z-5i}{iz-1}}$

$\textsf{The invariant points are obtained by solving}$

$\textsf{the following equation}$

$\mathsf{z=\dfrac{3z-5i}{iz-1}}$

$\mathsf{z(iz-1)=3z-5i}$

$\mathsf{iz^2-z=3z-5i}$

$\mathsf{iz^2-4z+5i=0}$

$\implies\mathsf{z^2+4i\,z+5=0}$

$\implies\mathsf{z^2+5i\,z-iz+5=0}$

$\implies\mathsf{z(z+5i)-i(z+5i)=0}$

$\implies\mathsf{(z+5i)\,(z-i)=0}$

$\implies\mathsf{z=-5i,i}$

$\therefore\textbf{The invariant points are -5i and i}$

Answer 2

Answer:

Transformation is :

W = (3z-5i)/(iz-1)​

∴To find "The invariant point of the transformation"

Solution:

$w = (3z-5i)/(iz-1)$

Take

$z=(3z-5i)/(iz-1)$

$z(iz-1) = (3z-5i)$

$iz^2-z = 3z-5i$

$iz^2-z-3z+5i=0$

$iz^2-4z+5i=0$

            Multiply by ( $i$ )

$i^2z-4iz+5i^2=0$

           ∵ $i^2 = -1$

 ⇒ $z^2+4iz+5 =0$

           we can write $4iz$ ↔ $5iz -iz$

 ⇒ $z^2+5iz-iz+5=0$

           we can write $+5$ ↔ $-5i^2$, Since $i^2 = -1$

 ⇒$z^2+5iz-iz-5i^2 =0$

$z(z+5i)-i(z+5i) =0$

$(z+5i)(z-i) =0$

$z=-5i,i$

∴ The invariant points are -5i, i

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