Question

compute the invariant points of the transformation w= -2z+4i/iz+1 ​

Answer 1

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

$\mathsf{w=\dfrac{-2z+4i}{iz+1}}$

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

$\mathsf{Invariant\;points\;of\;w=\dfrac{-2z+4i}{iz+1}}$

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

$\mathsf{Consider,}$

$\mathsf{w=\dfrac{-2z+4i}{iz+1}}$

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

$\mathsf{the\;equation}\;\;\bf\;z=\dfrac{-2z+4i}{iz+1}$

$\implies\mathsf{z(iz+1)=-2z+4i}$

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

$\implies\mathsf{iz^2+3z-4i=0}$

$\implies\mathsf{z^2-3i\,z-4=0}$

$\textsf{This can be written as,}$

$\implies\mathsf{z^2-3i\,z=4}$

$\implies\mathsf{z^2-3i\,z-\dfrac{9}{4}=4-\dfrac{9}{4}}$

$\implies\mathsf{\left(z-\dfrac{3i}{2}\right)^2=\dfrac{7}{4}}$

$\textsf{Taking square root on bothsides, we get}$

$\mathsf{z-\dfrac{3i}{2}=\pm\,\dfrac{\sqrt{7}}{2}}$

$\implies\mathsf{z=\dfrac{3i}{2}\pm\,\dfrac{\sqrt{7}}{2}}$

$\implies\boxed{\mathsf{z=\dfrac{3i\,\pm\,\sqrt{7}}{2}}}$

$\therefore\textbf{The invariant points are}\;\bf\,\dfrac{3i\,+\,\sqrt{7}}{2}\;\;\&\;\;\dfrac{3i\,-\,\sqrt{7}}{2}$

Answer 2

Concept: Invariant points means that the point on a graph than remains unchanged after the transformation applied on it.

Given: w= -2z+4i/iz+1 ​

Find: Find the invariant points of the transformation.

Solution: Let w=z

$w=\frac{-2z+4i}{iz+1}$

  = z(iz+1)=-2z+4i

  = i$z^{2}$+z=-2z+4i

  =  i$z^{2}$+z-4i+2z=0

  =  $z^{2}$-3iz-4=0

  =   $z^{2}$-3iz=4

   = $z^{2}-3iz-\frac{9}{4} =4-\frac{9}{4}$

   = $(z-\frac{3i}{2}) ^{2}=\frac{7}{4}$

Square root on both sides.

  $z-\frac{3i}{2}= \frac{\sqrt{7} }{2}$

   $z-\frac{3i}{2}$=±$\frac{\sqrt{7} }{2}$

  z= $\frac{3i}{2}$±$\frac{\sqrt{7} }{2}$

Final answer: The invariant points of the transformation are $\frac{3i}{2}$+$\frac{\sqrt{7} }{2}$ and $\frac{3i}{2}-\frac{\sqrt{7}}{2}$.

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