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Question Number 138765 by 676597498 last updated on 18/Apr/21

Commented by 676597498 last updated on 18/Apr/21

$$pleasehelp$$

Answered by physicstutes last updated on 18/Apr/21

$$\left(\mathrm{i}\right)T_1\mathrm{is}\mathrm{a}\mathrm{rotation}\mathrm{about}\mathrm{the}w\mathrm{plane}\mathrm{through}180°.$$$$T_2\mathrm{is}\mathrm{an}\mathrm{enlargment}\mathrm{scale}\mathrm{factor}3\mathrm{and}\mathrm{a}\mathrm{translation}\mathrm{of}(-1,2).$$$$\left(\mathrm{ii}\right)\mathrm{for}\mathrm{an}\mathrm{arbitiary}\mathrm{complex}\mathrm{number}z,f\left(z\right)=z\mathrm{if}\mathrm{it}\mathrm{is}\mathrm{invariant}$$$$\mathrm{then},z=3z-1+2i\Rightarrow -2z=-1+2i$$$$\mathrm{hence}\mathrm{invariate}\mathrm{complex}\mathrm{number}\mathrm{is}z=\frac{1}{2}-i$$$$\left(\mathrm{iii}\right)\mathrm{let}A\left(a\right)\mathrm{have}\mathrm{image}\overline{A}\left(\overline{a}\right)\mathrm{and}B\left(b\right)\mathrm{have}\mathrm{image}\overline{B}\left(\overline{b}\right)$$$$\mathrm{then},\overline{a}=2a-1+2i.........\left(i\right)$$$$\mathrm{and}\overline{b}=2b-1+2i............\left(ii\right)$$$$\left(ii\right)-\left(i\right)\Rightarrow \overline{b}-\overline{a}=2(b-a)$$$$\Rightarrow \mid \overline{b}-\overline{a}\mid =2\mid b-a\mid$$$$\mathrm{hence}\mid \overline{B}\overline{A}\mid =2\mid BA\mid \mathrm{and}\mathrm{so}{T}_2\mathrm{is}\mathrm{a}\mathrm{similarity}\mathrm{transformation}\mathrm{or}\mathrm{a}\mathit{s}\mathit{i}\mathit{m}\mathit{u}\mathit{l}\mathit{t}\mathit{i}\mathit{t}\mathit{u}\mathit{d}\mathit{e}$$$$\mathrm{with}\mathrm{radius}2.$$

Answered by TheSupreme last updated on 18/Apr/21

$$z=\{Re\left(z\right),Im\left(z\right)\}$$$$w=-\{Re\left(z\right),-Im\left(z\right)\}=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]{\left\{z\right\}}^T$$$$T_{1}isasimmetryonreal-axis$$$$w=3z-1+2i=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]{\left\{z\right\}}^T+\left[\begin{array}{c}-1\\ 2i\end{array}\right]$$$$T_{2}isascalewithcenter(\frac{-1}{3},\frac{2i}{3})>$$

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