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Question Number 88479 by Ar Brandon last updated on 10/Apr/20
Consider the transformation f of the plane with all points  M wity affix z mapped to the point M ′ with affix z ′  such that z ′=−((√3)+i)z−1+i(1+(√3))  1) Given M_0  the point z_0 =((√3)/4)+(3/4)i  calculate AM_0  and deduce the angle in radians  (Taking A as the center of the transformation)  2) Consider the progression with points(M_n )_(n≥0)  defined by  f(M_n )=M_(n+1)   a∙ Show by recurrence that ∀n∈N z_n =2^n e^(ln((7π)/6))  (z_(0 ) −i)  Find AM_n  then determine the smallest natural number, n, such that  AM_n ≥10^2
$${Consider}\:{the}\:{transformation}\:\boldsymbol{{f}}\:{of}\:{the}\:{plane}\:{with}\:{all}\:{points} \\ $$$$\boldsymbol{{M}}\:{wity}\:{affix}\:\boldsymbol{{z}}\:{mapped}\:{to}\:{the}\:{point}\:\boldsymbol{{M}}\:'\:{with}\:{affix}\:\boldsymbol{{z}}\:' \\ $$$${such}\:{that}\:\boldsymbol{{z}}\:'=−\left(\sqrt{\mathrm{3}}+{i}\right){z}−\mathrm{1}+{i}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\left.\mathrm{1}\right)\:{Given}\:\boldsymbol{{M}}_{\mathrm{0}} \:{the}\:{point}\:\boldsymbol{{z}}_{\mathrm{0}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}{i} \\ $$$${calculate}\:\boldsymbol{{AM}}_{\mathrm{0}} \:{and}\:{deduce}\:{the}\:{angle}\:{in}\:{radians} \\ $$$$\left({Taking}\:\boldsymbol{{A}}\:{as}\:{the}\:{center}\:{of}\:{the}\:{transformation}\right) \\ $$$$\left.\mathrm{2}\right)\:{Consider}\:{the}\:{progression}\:{with}\:{points}\left(\boldsymbol{{M}}_{\boldsymbol{{n}}} \right)_{\boldsymbol{{n}}\geqslant\mathrm{0}} \:{defined}\:{by} \\ $$$${f}\left({M}_{{n}} \right)={M}_{{n}+\mathrm{1}} \\ $$$${a}\centerdot\:{Show}\:{by}\:{recurrence}\:{that}\:\forall{n}\in\mathbb{N}\:\boldsymbol{{z}}_{\boldsymbol{{n}}} =\mathrm{2}^{{n}} {e}^{{ln}\frac{\mathrm{7}\pi}{\mathrm{6}}} \:\left({z}_{\mathrm{0}\:} −{i}\right) \\ $$$${Find}\:{AM}_{{n}} \:{then}\:{determine}\:{the}\:{smallest}\:{natural}\:{number},\:{n},\:{such}\:{that} \\ $$$${AM}_{{n}} \geqslant\mathrm{10}^{\mathrm{2}} \\ $$

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