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Let $V_1$ be the anti-clockwise rotation in the plane about origin with $\theta$ angle and $V_2$ be the anti-clockwise rotation in the plane about (2,0) by $\theta$ angle. finding its center of rotation of composition $V_2(V_1)$

around origin i have a transformation for $V_1$ that is $f_1(z)=e^{\iota\theta} z$ similarly for $V_2$ i have $f_2(z)=e^{\iota\theta} (z-2)+2$ then $f_2f_1(z)=e^{\iota\theta} ( f_1(z)-2)+2=e^{\iota\theta} (e^{\iota\theta}z-2)+2$. now how to find center from this without using matrices.

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  • $\begingroup$ there is a mistake in your computations for $f_2\circ f_1$. also, what makes you think that the transformation $V_2\circ V_1$ will be a rotation? $\endgroup$
    – AnCar
    Commented Jun 15 at 13:39
  • $\begingroup$ i have edited it, it look likes transformation of $2\theta$, $\endgroup$
    – Ricci Ten
    Commented Jun 15 at 13:45
  • $\begingroup$ look for example at the points $(0,0)$ and $(1,0)$. does there distance remain $1$ after applying $V_2\circ V_1$? $\endgroup$
    – AnCar
    Commented Jun 15 at 13:48
  • $\begingroup$ if i know about center of rotation from there distance should preserve $\endgroup$
    – Ricci Ten
    Commented Jun 15 at 13:54
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    $\begingroup$ Solve $f_2(f_1(z))=z$ for $z$. $\endgroup$ Commented Jun 15 at 14:01

1 Answer 1

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$ f_2 (f_1 (z)) = e^{i \theta } ( e^{i \theta} z - 2 ) + 2 = e^{i 2 \theta} z - 2 e^{i \theta } + 2 $

And this is a rotation by $2 \theta$. To find the center, take the general form of rotation by $ 2 \theta $ about $z_0$:

$ g(z) = e^{i 2 \theta} (z - z_0 ) + z_0 $

And compare this form to the earlier equation. We deduce that

$ z_0 ( 1 - e^{i 2 \theta } ) = 2 ( 1 - e^{i \theta} ) $

Therefore, the center of rotation is given by

$ z_0 = \dfrac{ 2 (1 - e^{i \theta} ) }{ 1 - e^{i 2 \theta} } = \dfrac{2}{1 + e^{i \theta} } $

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    $\begingroup$ yes center is $(1,-\tan{\frac{\theta}{2}})$. $\endgroup$
    – Ricci Ten
    Commented Jun 15 at 14:30

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