Math, asked by muralitharanks13, 5 months ago

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

Answers

Answered by MaheswariS
2

\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}

Answered by KABIIL
0

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 4iz5iz -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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