1. **Problem statement:** We consider a rectangle OABE in an oriented plane with OA = 2 and angle between vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ equal to $\frac{\pi}{3}$. The circle (C) has diameter [OB] and center W. The similitude S has center O, ratio $\sqrt{3}$, and angle $\frac{\pi}{3}$. ### A- 1) Let $A'$ be on the ray [OB) such that $OA' = 2\sqrt{3}$. Prove $A'$ is image of A under S. - Since S is a direct similitude with center O, ratio $\sqrt{3}$, and angle $\frac{\pi}{3}$, the image of A is given by: $$OA' = \sqrt{3} \times OA = \sqrt{3} \times 2 = 2\sqrt{3}$$ - The angle between $\overrightarrow{OA}$ and $\overrightarrow{OA'}$ is $\frac{\pi}{3}$ by definition of S. - Since $A'$ lies on [OB) and satisfies these conditions, $A' = S(A)$. 2a) Verify triangle OAW is equilateral. - W is center of circle with diameter [OB], so W is midpoint of OB. - Length $OB = \sqrt{OA^2 + AB^2}$. Since OABE is rectangle, $AB = OE = 2\sqrt{3}$ (from given z_E = 2\sqrt{3}i). - Calculate $OB = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$. - W is midpoint of OB, so $OW = WB = 2$. - Triangle OAW has sides: - $OA = 2$ - $AW = ?$ (to be calculated) - $OW = 2$ - Vector $\overrightarrow{AW} = \overrightarrow{OW} - \overrightarrow{OA} = W - A$. - Using complex affixes: $z_A = 2$, $z_B = 2 + 2\sqrt{3}i$, so $z_W = \frac{z_O + z_B}{2} = \frac{0 + (2 + 2\sqrt{3}i)}{2} = 1 + \sqrt{3}i$. - Length $AW = |z_W - z_A| = |(1 + \sqrt{3}i) - 2| = | -1 + \sqrt{3}i| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. - All sides $OA = OW = AW = 2$, so triangle OAW is equilateral. 2b) Determine image under S of triangle OAW. - S multiplies lengths by $\sqrt{3}$ and rotates by $\frac{\pi}{3}$. - Since O is center, image of O is O. - Image of A is $A'$ with $OA' = 2\sqrt{3}$ and angle $\frac{\pi}{3}$ from OA. - Image of W is $W' = S(W)$. - Since triangle OAW is equilateral, its image is also equilateral with side lengths multiplied by $\sqrt{3}$. 2c) Construct circle (C'), image of (C) under S. - Circle (C) has center W and radius $\frac{OB}{2} = 2$. - Under S, center W maps to W', radius multiplies by $\sqrt{3}$, so radius of (C') is $2 \times \sqrt{3} = 2\sqrt{3}$. - Thus, (C') is circle with center W' and radius $2\sqrt{3}$. ### B- 1) Write complex form of S. - S is direct similitude with center O, ratio $\sqrt{3}$, angle $\frac{\pi}{3}$. - Complex form: $z' = \sqrt{3} e^{i \pi/3} z = \sqrt{3} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) z = \sqrt{3} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) z = \frac{\sqrt{3}}{2} z + i \frac{3}{2} z$. 2) Find affixes of W and W' = S(W). - $z_W = 1 + \sqrt{3} i$ (midpoint of OB). - $z_{W'} = S(z_W) = \sqrt{3} e^{i \pi/3} (1 + \sqrt{3} i)$. - Calculate: $$e^{i \pi/3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$$ $$z_{W'} = \sqrt{3} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) (1 + \sqrt{3} i)$$ - Multiply inside: $$(1 + \sqrt{3} i) \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{1}{2} + i \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} i + i^2 \frac{3}{2} = \frac{1}{2} + i \frac{\sqrt{3}}{2} + i \frac{\sqrt{3}}{2} - \frac{3}{2}$$ $$= \left( \frac{1}{2} - \frac{3}{2} \right) + i \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right) = -1 + i \sqrt{3}$$ - Then: $$z_{W'} = \sqrt{3} (-1 + i \sqrt{3}) = -\sqrt{3} + 3 i$$ 3a) Show $f(z) = i z + 4 + 2 i \sqrt{3}$ is a rotation, find angle and center H. - A transformation $f(z) = e^{i \theta} z + b$ is a rotation if $|e^{i \theta}| = 1$. - Here, $e^{i \theta} = i = e^{i \pi/2}$, so angle $\theta = \frac{\pi}{2}$. - Center H satisfies: $$f(H) = H \Rightarrow i H + 4 + 2 i \sqrt{3} = H$$ $$i H - H = -4 - 2 i \sqrt{3}$$ $$(i - 1) H = -4 - 2 i \sqrt{3}$$ $$H = \frac{-4 - 2 i \sqrt{3}}{i - 1}$$ - Multiply numerator and denominator by conjugate $1 + i$: $$H = \frac{(-4 - 2 i \sqrt{3})(1 + i)}{(i - 1)(1 + i)}$$ - Denominator: $$(i - 1)(1 + i) = i + i^2 - 1 - i = i - 1 - 1 - i = -2$$ - Numerator: $$(-4)(1) + (-4)(i) + (-2 i \sqrt{3})(1) + (-2 i \sqrt{3})(i) = -4 - 4 i - 2 i \sqrt{3} - 2 i^2 \sqrt{3}$$ $$= -4 - 4 i - 2 i \sqrt{3} + 2 \sqrt{3}$$ $$= (-4 + 2 \sqrt{3}) + (-4 i - 2 i \sqrt{3})$$ - So: $$H = \frac{(-4 + 2 \sqrt{3}) + i(-4 - 2 \sqrt{3})}{-2} = 2 - \sqrt{3} + i (2 + \sqrt{3})$$ 3b) Verify $f(W') = W$ and find $f \circ S (W)$. - $f(W') = i z_{W'} + 4 + 2 i \sqrt{3} = i (-\sqrt{3} + 3 i) + 4 + 2 i \sqrt{3}$ - Calculate: $$i (-\sqrt{3}) + i (3 i) + 4 + 2 i \sqrt{3} = - i \sqrt{3} + 3 i^2 + 4 + 2 i \sqrt{3} = - i \sqrt{3} - 3 + 4 + 2 i \sqrt{3}$$ $$= (4 - 3) + (- i \sqrt{3} + 2 i \sqrt{3}) = 1 + i \sqrt{3} = z_W$$ - So $f(W') = W$. - Then $f \circ S (W) = f(W') = W$. 3c) Determine nature and characteristic elements of $f \circ S$. - $f \circ S (z) = f(S(z)) = i (\sqrt{3} e^{i \pi/3} z) + 4 + 2 i \sqrt{3} = \sqrt{3} i e^{i \pi/3} z + 4 + 2 i \sqrt{3}$. - Simplify rotation factor: $$i e^{i \pi/3} = e^{i \pi/2} e^{i \pi/3} = e^{i (\pi/2 + \pi/3)} = e^{i (5\pi/6)}$$ - So: $$f \circ S (z) = \sqrt{3} e^{i 5\pi/6} z + 4 + 2 i \sqrt{3}$$ - This is a similitude with ratio $\sqrt{3}$ and rotation angle $\frac{5\pi}{6}$. - The translation part means it is a rotation + homothety (direct similitude) with center to be found by solving: $$f \circ S (H) = H$$ $$\sqrt{3} e^{i 5\pi/6} H + 4 + 2 i \sqrt{3} = H$$ $$(\sqrt{3} e^{i 5\pi/6} - 1) H = -4 - 2 i \sqrt{3}$$ - The nature is a direct similitude with center H, ratio $\sqrt{3}$, and angle $\frac{5\pi}{6}$. **Final answers:** - A' is image of A under S. - Triangle OAW is equilateral. - Image of OAW under S is equilateral with side lengths multiplied by $\sqrt{3}$. - Circle (C') has center W' and radius $2 \sqrt{3}$. - Complex form of S: $z' = \sqrt{3} e^{i \pi/3} z$. - $z_W = 1 + \sqrt{3} i$, $z_{W'} = -\sqrt{3} + 3 i$. - $f$ is rotation of angle $\frac{\pi}{2}$ with center $H = 2 - \sqrt{3} + i (2 + \sqrt{3})$. - $f(W') = W$, so $f \circ S (W) = W$. - $f \circ S$ is direct similitude with ratio $\sqrt{3}$, angle $\frac{5\pi}{6}$, center H solving above equation.