If z = fracsqrt32 + fraci2, i = sqrt-1$z = \frac{\sqrt{3}}{2} + \frac{i}{2}, i = \sqrt{-1}$, then (z^201 - i)^8$(z^{201} - i)^8$ is equal to

### Related Formula x^2 + x + 1 = 0 Rightarrow x = omega, omega^2$$x^2 + x + 1 = 0 \Rightarrow x = \omega, \omega^2$$ Properties of cube roots of unity: omega^3 = 1$\omega^3 = 1$ and 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$. ### Core Logic Let alpha = omega$\alpha = \omega$. Then frac1x = frac1omega = omega^2$\frac{1}{x} = \frac{1}{\omega} = \omega^2$. The series is sum_k=1^25 (omega^k + omega^2k)^4$\sum_{k=1}^{25} (\omega^k + \omega^{2k})^4$. Evaluate the term T_k = (omega^k + omega^2k)^4$T_k = (\omega^k + \omega^{2k})^4$ based on the modulo of k$k$ with 3. ### Step 1: Cyclic Evaluation Case 1: k = 3m$k = 3m$ (multiples of 3) T_3m = (omega^3m + omega^6m)^4 = (1 + 1)^4 = 2^4 = 16$$T_{3m} = (\omega^{3m} + \omega^{6m})^4 = (1 + 1)^4 = 2^4 = 16$$ There are 8$8$ such multiples up to 25 (3, 6, 9, dots, 24$3, 6, 9, \dots, 24$). Case 2: k neq 3m$k \neq 3m$ (non-multiples of 3) For k = 1, 2, 4, 5, dots$k = 1, 2, 4, 5, \dots$ omega^k + omega^2k$\omega^k + \omega^{2k}$ will always be omega + omega^2$\omega + \omega^2$ or omega^2 + omega$\omega^2 + \omega$. Since 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$, we have omega + omega^2 = -1$\omega + \omega^2 = -1$. T_k neq 3m = (-1)^4 = 1$$T_{k \neq 3m} = (-1)^4 = 1$$ There are 25 - 8 = 17$25 - 8 = 17$ such non-multiples up to 25. ### Step 2: Total Sum textSum = 17 times 1 + 8 times 16$$\text{Sum} = 17 \times 1 + 8 \times 16$$ = 17 + 128 = 145$$= 17 + 128 = 145$$ ### Pattern Recognition Powers of x+1/x$x+1/x$ when x$x$ solves x^2 pm x + 1 = 0$x^2 \pm x + 1 = 0$ perfectly orbit around periods of 3 or 6. Isolate the 3m$3m$ resonant beats (which hit pure scalars like 1+1=2$1+1=2$) versus the out-of-phase beats (which collapse to -1$-1$ or 1$1$ via basic omega$\omega$ identities). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Related Formula textFor maximum argument of z text on a circle, the ray from origin is tangent to the circle.$$\text{For maximum argument of } z \text{ on a circle, the ray from origin is tangent to the circle.}$$ textIf |z-a| leq r, text maximum argument implies sintheta = fracr|a| text and coordinates are x=acos^2theta, y=asinthetacostheta$$\text{If } |z-a| \leq r, \text{ maximum argument implies } \sin\theta = \frac{r}{|a|} \text{ and coordinates are } x=a\cos^2\theta, y=a\sin\theta\cos\theta$$ ### Core Logic

The condition |z-5|leq 3$|z-5|\leq 3$ represents a disk centered at (5,0)$(5,0)$ with radius 3$3$. To maximize the principal argument theta$\theta$, the ray from origin must touch the circle in the first quadrant. The tangent, origin, and center form a right-angled triangle. ### Step 1: Locate the Point z From geometry, hypotenuse c = 5$c = 5$, opposite (radius) r = 3$r = 3$. The adjacent side (length of tangent) is sqrt5^2 - 3^2 = 4$\sqrt{5^2 - 3^2} = 4$. The angle of tangency theta$\theta$ satisfies sintheta = frac35$\sin\theta = \frac{3}{5}$ and costheta = frac45$\cos\theta = \frac{4}{5}$. The point P(z)$P(z)$ lies on the circle and the tangent ray: z equiv (4costheta, 4sintheta) = left(4left(frac45right), 4left(frac35right)right) = left(frac165, frac125right)$$z \equiv (4\cos\theta, 4\sin\theta) = \left(4\left(\frac{4}{5}\right), 4\left(\frac{3}{5}\right)\right) = \left(\frac{16}{5}, \frac{12}{5}\right)$$ z = frac165 + frac125i$$z = \frac{16}{5} + \frac{12}{5}i$$ ### Step 2: Evaluate the Target Expression We need 34left|frac5z - 125iz + 16right|^2$34\left|\frac{5z - 12}{5iz + 16}\right|^2$. Substitute 5z = 16 + 12i$5z = 16 + 12i$: Numerator: 5z - 12 = 16 + 12i - 12 = 4 + 12i$5z - 12 = 16 + 12i - 12 = 4 + 12i$ Denominator: 5iz + 16 = i(16 + 12i) + 16 = 16i - 12 + 16 = 4 + 16i$5iz + 16 = i(16 + 12i) + 16 = 16i - 12 + 16 = 4 + 16i$ Expression: 34 left|frac4 + 12i4 + 16iright|^2 = 34 frac|4 + 12i|^2|4 + 16i|^2 = 34 frac4^2 + 12^24^2 + 16^2$$34 \left|\frac{4 + 12i}{4 + 16i}\right|^2 = 34 \frac{|4 + 12i|^2}{|4 + 16i|^2} = 34 \frac{4^2 + 12^2}{4^2 + 16^2}$$ = 34 left(frac16 + 14416 + 256right) = 34 left(frac160272right)$$= 34 \left(\frac{16 + 144}{16 + 256}\right) = 34 \left(\frac{160}{272}\right)$$ = frac34 times 160272 = frac5440272 = 20$$= \frac{34 \times 160}{272} = \frac{5440}{272} = 20$$ ### Pattern Recognition For maximum arg(z)$\arg(z)$ on |z-c| = r$|z-c| = r$ where c$c$ is real, z$z$ coordinates are given by geometric projection: z = sqrtc^2-r^2(costheta + isintheta)$z = \sqrt{c^2-r^2}(\cos\theta + i\sin\theta)$ where sintheta = r/c$\sin\theta = r/c$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Related Formula alpha = omega, quad beta = omega^2$$\alpha = \omega, \quad \beta = \omega^2$$ 1 + omega + omega^2 = 0 implies omega^3 = 1$$1 + \omega + \omega^2 = 0 \implies \omega^3 = 1$$ ### Core Logic Substitute omega$\omega$ and omega^2$\omega^2$ into the terms: Let T_1 = 7 - 7omega + 9omega^2$T_1 = 7 - 7\omega + 9\omega^2$ Let T_2 = 9 + 7omega - 7omega^2$T_2 = 9 + 7\omega - 7\omega^2$ Let T_3 = -7 + 9omega + 7omega^2$T_3 = -7 + 9\omega + 7\omega^2$ Let T_4 = 14 + 7omega + 7omega^2$T_4 = 14 + 7\omega + 7\omega^2$ ### Step 1: Simplify Each Term Using Properties of Omega Notice relationships between the terms by factoring powers of omega$\omega$: T_1 = omega cdot omega^2(7 - 7omega + 9omega^2) text wait, an easier path is factoring$$T_1 = \omega \cdot \omega^2(7 - 7\omega + 9\omega^2) \text{ wait, an easier path is factoring}$$ Observe that: omega cdot T_2 = omega(9 + 7omega - 7omega^2) = 9omega + 7omega^2 - 7 = T_3$$\omega \cdot T_2 = \omega(9 + 7\omega - 7\omega^2) = 9\omega + 7\omega^2 - 7 = T_3$$ omega^2 cdot T_2 = omega^2(9 + 7omega - 7omega^2) = 9omega^2 + 7 - 7omega = T_1$$\omega^2 \cdot T_2 = \omega^2(9 + 7\omega - 7\omega^2) = 9\omega^2 + 7 - 7\omega = T_1$$ So, T_1^20 + T_2^20 + T_3^20 = (omega^2 T_2)^20 + T_2^20 + (omega T_2)^20$T_1^{20} + T_2^{20} + T_3^{20} = (\omega^2 T_2)^{20} + T_2^{20} + (\omega T_2)^{20}$ = omega^40 T_2^20 + T_2^20 + omega^20 T_2^20 = T_2^20 (1 + omega^40 + omega^20)$$= \omega^{40} T_2^{20} + T_2^{20} + \omega^{20} T_2^{20} = T_2^{20} (1 + \omega^{40} + \omega^{20})$$ Since omega^3 = 1$\omega^3 = 1$, we have omega^40 = omega$\omega^{40} = \omega$ and omega^20 = omega^2$\omega^{20} = \omega^2$. Thus, T_2^20 (1 + omega + omega^2) = T_2^20 (0) = 0$T_2^{20} (1 + \omega + \omega^2) = T_2^{20} (0) = 0$. ### Step 2: Evaluating the Non-Zero Term The only remaining term is T_4^20$T_4^{20}$: T_4 = 14 + 7(omega + omega^2)$$T_4 = 14 + 7(\omega + \omega^2)$$ Since omega + omega^2 = -1$\omega + \omega^2 = -1$, T_4 = 14 + 7(-1) = 14 - 7 = 7$$T_4 = 14 + 7(-1) = 14 - 7 = 7$$ So the entire expression sum is 0 + 7^20 = 7^20$0 + 7^{20} = 7^{20}$. ### Step 3: Finding m Given that the sum equals m^10$m^{10}$: m^10 = 7^20 = (7^2)^10 = 49^10$$m^{10} = 7^{20} = (7^2)^{10} = 49^{10}$$ Hence, m = 49$m = 49$.

### Pattern Recognition Complex expressions with omega$\omega$ raised to large powers almost always exploit rotational symmetry. If coefficients are permuted circularly (A, B, C) to (C, A, B) to (B, C, A)$(A, B, C) \to (C, A, B) \to (B, C, A)$, multiplying by omega$\omega$ maps them to one another, meaning their sum of 3n$3n$ or symmetric powers identically nullifies. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Related Formula For z = x + iy$z = x + iy$, overlinez = x - iy$\overline{z} = x - iy$ and |z|^2 = x^2 + y^2$|z|^2 = x^2 + y^2$. ### Core Logic Substitute z = x + iy$z = x + iy$ into 4z^2 + overlinez = 0$4z^2 + \overline{z} = 0$: 4(x^2 - y^2 + 2ixy) + x - iy = 0$$4(x^2 - y^2 + 2ixy) + x - iy = 0$$ Separating real and imaginary parts: 4x^2 - 4y^2 + x = 0 quad textand quad y(8x - 1) = 0$$4x^2 - 4y^2 + x = 0 \quad \text{and} \quad y(8x - 1) = 0$$ Case 1: y = 0 implies 4x^2 + x = 0 implies x = 0$y = 0 \implies 4x^2 + x = 0 \implies x = 0$ or x = -1/4$x = -1/4$. Solutions: z_1 = 0 implies |z_1|^2 = 0$z_1 = 0 \implies |z_1|^2 = 0$, z_2 = -1/4 implies |z_2|^2 = 1/16$z_2 = -1/4 \implies |z_2|^2 = 1/16$. Case 2: x = 1/8 implies 4(1/64) - 4y^2 + 1/8 = 0 implies 4y^2 = 3/16 implies y = pm fracsqrt38$x = 1/8 \implies 4(1/64) - 4y^2 + 1/8 = 0 \implies 4y^2 = 3/16 \implies y = \pm \frac{\sqrt{3}}{8}$. Solutions: z_3, z_4 = frac18 pm i fracsqrt38 implies |z_3|^2 = |z_4|^2 = frac164 + frac364 = frac116$z_3, z_4 = \frac{1}{8} \pm i \frac{\sqrt{3}}{8} \implies |z_3|^2 = |z_4|^2 = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$. ### Step 1: Sum of Modulus Squared sum_z in S |z|^2 = 0 + frac116 + frac116 + frac116 = frac316$$\sum_{z \in S} |z|^2 = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$$ ### Pattern Recognition Separate complex equation into real and imaginary components to systematically find all roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Core Logic Simplify the condition for set S$S$: 3 leq left|2left(z - frac32(1 + i)right)right| leq 7$$3 \leq \left|2\left(z - \frac{3}{2}(1 + i)\right)\right| \leq 7$$ frac32 leq left|z - frac32(1 + i)right| leq frac72$$\frac{3}{2} \leq \left|z - \frac{3}{2}(1 + i)\right| \leq \frac{7}{2}$$ This represents an annular region bounded by two concentric circles centered at C = frac32 + frac32i$C = \frac{3}{2} + \frac{3}{2}i$ with radii r_1 = frac32$r_1 = \frac{3}{2}$ and r_2 = frac72$r_2 = \frac{7}{2}$.

### Step 1: Identify the Target Point We need to minimize left| z - left( -frac52 - frac32i right) right|$\left| z - \left( -\frac{5}{2} - \frac{3}{2}i \right) \right|$. Let P$P$ be the point -frac52 - frac32i$-\frac{5}{2} - \frac{3}{2}i$. The expression represents the distance from point P$P$ to a point z$z$ in the set S$S$. ### Step 2: Distance from Center to P Calculate the distance PC$PC$ between the center of the circles Cleft(frac32, frac32right)$C\left(\frac{3}{2}, \frac{3}{2}\right)$ and the point Pleft(-frac52, -frac32right)$P\left(-\frac{5}{2}, -\frac{3}{2}\right)$: PC = sqrtleft( frac32 - left(-frac52right) right)^2 + left( frac32 - left(-frac32right) right)^2$$PC = \sqrt{\left( \frac{3}{2} - \left(-\frac{5}{2}\right) \right)^2 + \left( \frac{3}{2} - \left(-\frac{3}{2}\right) \right)^2}$$ PC = sqrtleft( frac82 right)^2 + left( frac62 right)^2 = sqrt4^2 + 3^2 = sqrt16 + 9 = sqrt25 = 5$$PC = \sqrt{\left( \frac{8}{2} \right)^2 + \left( \frac{6}{2} \right)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ Since 5 > frac72$5 > \frac{7}{2}$, the point P$P$ lies outside the outer circle. ### Step 3: Minimum Distance Calculation The shortest distance from an external point to an annular region is the distance to the outer boundary along the line connecting the point to the center. min_z in S |z - P| = PC - r_textouter$$\min_{z \in S} |z - P| = PC - r_{\text{outer}}$$ textMinimum Distance = 5 - frac72 = frac10 - 72 = frac32$$\text{Minimum Distance} = 5 - \frac{7}{2} = \frac{10 - 7}{2} = \frac{3}{2}$$ ### Pattern Recognition Transforming |az - b|$|az - b|$ by factoring out a$a$ immediately reveals the true geometric center. Shortest distance to any ring/circle from an external point is always collinear with the center: d - R_outer$d - R_{outer}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations