Solution & Explanation
### Related Formula
textFor a quadratic equation A x^2 + B x + C = 0 text to have two positive real roots:$$\text{For a quadratic equation } A x^2 + B x + C = 0 \text{ to have two positive real roots:}$$
text1. Real roots: D = B^2 - 4AC ge 0$$\text{1. Real roots: } D = B^2 - 4AC \ge 0$$
text2. Sum of roots: -fracBA > 0$$\text{2. Sum of roots: } -\frac{B}{A} > 0$$
text3. Product of roots: fracCA > 0$$\text{3. Product of roots: } \frac{C}{A} > 0$$
### Core Logic
We write down three systems of inequalities based on real and positive root conditions, find their intersection, and map the boundaries to solve for the parameters.
### Step 1: Apply the discriminant condition (Real roots)
For real roots, the discriminant D ge 0$D \ge 0$:
D = left[ 2(a - 3) right]^2 - 4(1 - a)(9) ge 0$$D = \left[ 2(a - 3) \right]^2 - 4(1 - a)(9) \ge 0$$
4(a^2 - 6a + 9) - 36(1 - a) ge 0$$4(a^2 - 6a + 9) - 36(1 - a) \ge 0$$
(a^2 - 6a + 9) - 9(1 - a) ge 0$$(a^2 - 6a + 9) - 9(1 - a) \ge 0$$
a^2 - 6a + 9 - 9 + 9 a ge 0$$a^2 - 6a + 9 - 9 + 9 a \ge 0$$
a^2 + 3a ge 0 implies a(a + 3) ge 0$$a^2 + 3a \ge 0 \implies a(a + 3) \ge 0$$
Thus, the interval is:
a in (-infty, -3] cup [0, infty) quad text--- (1)$$a \in (-\infty, -3] \cup [0, \infty) \quad \text{--- (1)}$$
### Step 2: Apply the sum of roots condition (Positive sum)
For positive roots, the sum of roots must be positive:
-fracBA = frac-2(a - 3)1 - a = frac2(a - 3)a - 1 > 0$$-\frac{B}{A} = \frac{-2(a - 3)}{1 - a} = \frac{2(a - 3)}{a - 1} > 0$$
Using the wavy curve method for fraca-3a-1 > 0$\frac{a-3}{a-1} > 0$:
a in (-infty, 1) cup (3, infty) quad text--- (2)$$a \in (-\infty, 1) \cup (3, \infty) \quad \text{--- (2)}$$
### Step 3: Apply the product of roots condition (Positive product)
For positive roots, the product of roots must be positive:
fracCA = frac91 - a > 0 implies 1 - a > 0 implies a < 1$$\frac{C}{A} = \frac{9}{1 - a} > 0 \implies 1 - a > 0 \implies a < 1$$
Thus, the interval is:
a in (-infty, 1) quad text--- (3)$$a \in (-\infty, 1) \quad \text{--- (3)}$$
### Step 4: Find the intersection of all conditions
Intersecting equations (1), (2), and (3):
- First, intersect (2) and (3):
( (-infty, 1) cup (3, infty) ) cap (-infty, 1) = (-infty, 1)$$( (-\infty, 1) \cup (3, \infty) ) \cap (-\infty, 1) = (-\infty, 1)$$
- Next, intersect with (1):
( (-infty, -3] cup [0, infty) ) cap (-infty, 1) = (-infty, -3] cup [0, 1)$$( (-\infty, -3] \cup [0, \infty) ) \cap (-\infty, 1) = (-\infty, -3] \cup [0, 1)$$
Comparing this with (-infty, -alpha] cup [beta, gamma)$(-\infty, -\alpha] \cup [\beta, \gamma)$:
- alpha = 3$\alpha = 3$
- beta = 0$\beta = 0$
- gamma = 1$\gamma = 1$
Now calculate the target sum:
2alpha + beta + gamma = 2(3) + 0 + 1 = 7$$2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 7$$
### Pattern Recognition
Location of roots: When both roots are positive, checking sum and product signs along with D ge 0$D \ge 0$ is the standard and fastest set of inequalities, avoiding complex vertex projections.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Practical Chemistry
More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 3
Q55
jee_main_2025_04_april_evening
Properties of Complex Numbers
Let the product of omega_1 = (8 + mathrmi) sin theta + (7 + 4 mathrmi) cos theta$\omega_1 = (8 + \mathrm{i}) \sin \theta + (7 + 4 \mathrm{i}) \cos \theta$ and omega_2 = (1 + 8 mathrmi) sin theta + (4 + 7 mathrmi) cos theta$\omega_2 = (1 + 8 \mathrm{i}) \sin \theta + (4 + 7 \mathrm{i}) \cos \theta$ be alpha +mathrmibeta$\alpha +\mathrm{i}\beta$, mathrmi = sqrt-1$\mathrm{i} = \sqrt{-1}$. Let p and q be the maximum and the minimum values of alpha +beta$\alpha +\beta$ respectively.
- A. 140$140$
- B. 130$130$
- C. 160$160$
- D. 150$150$
Solution
### Core Logic
Let's expand the terms by grouping real and imaginary parts explicitly:
omega_1 = (8sintheta + 7costheta) + i(sintheta + 4costheta)$$\omega_1 = (8\sin\theta + 7\cos\theta) + i(\sin\theta + 4\cos\theta)$$
omega_2 = (sintheta + 4costheta) + i(8sintheta + 7costheta)$$\omega_2 = (\sin\theta + 4\cos\theta) + i(8\sin\theta + 7\cos\theta)$$
Notice that if we let u = 8sintheta + 7costheta$u = 8\sin\theta + 7\cos\theta$ and v = sintheta + 4costheta$v = \sin\theta + 4\cos\theta$, then:
omega_1 = u + iv quad textand quad omega_2 = v + iu$$\omega_1 = u + iv \quad \text{and} \quad \omega_2 = v + iu$$
### Step 1: Calculating the Product
Multiplying omega_1$\omega_1$ and \omega_2:
omega_1omega_2 = (u + iv)(v + iu) = uv + iu^2 + iv^2 - uv = i(u^2 + v^2)$$\omega_1\omega_2 = (u + iv)(v + iu) = uv + iu^2 + iv^2 - uv = i(u^2 + v^2)$$
Since the product is given as alpha + ibeta$\alpha + i\beta$:
alpha = 0$\alpha = 0$
beta = u^2 + v^2 = (8sintheta + 7costheta)^2 + (sintheta + 4costheta)^2$$\beta = u^2 + v^2 = (8\sin\theta + 7\cos\theta)^2 + (\sin\theta + 4\cos\theta)^2$$
### Step 2: Simplifying the expression for alpha + beta
Expanding the terms for beta$\beta$:
beta = (64sin^2theta + 49cos^2theta + 112sinthetacostheta) + (sin^2theta + 16cos^2theta + 8sinthetacostheta)$$\beta = (64\sin^2\theta + 49\cos^2\theta + 112\sin\theta\cos\theta) + (\sin^2\theta + 16\cos^2\theta + 8\sin\theta\cos\theta)$$
alpha + beta = 0 + beta = 65sin^2theta + 65cos^2theta + 120sinthetacostheta$$\alpha + \beta = 0 + \beta = 65\sin^2\theta + 65\cos^2\theta + 120\sin\theta\cos\theta$$
Using the identity sin^2theta + cos^2theta = 1$\sin^2\theta + \cos^2\theta = 1$ and 2sinthetacostheta = sin 2theta$2\sin\theta\cos\theta = \sin 2\theta$:
alpha + beta = 65 + 60sin 2theta$$\alpha + \beta = 65 + 60\sin 2\theta$$
### Step 3: Max and Min Extrema Analysis
Since -1 le sin 2theta le 1$-1 \le \sin 2\theta \le 1$:
textMaximum value p = 65 + 60(1) = 125$$\text{Maximum value } p = 65 + 60(1) = 125$$
textMinimum value q = 65 + 60(-1) = 5$$\text{Minimum value } q = 65 + 60(-1) = 5$$
Sum of maximum and minimum bounds equals:
p + q = 125 + 5 = 130$$p + q = 125 + 5 = 130$$
### Pattern Recognition
Observe the symmetric structure in complex variables: (u+iv)$(u+iv)$ and (v+iu)$(v+iu)$. Their product structurally completely cancels out the real component, saving you from a highly messy component expansion.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers
Class 11 Mathematics: Trigonometric Functions
Q71
jee_main_2025_04_april_evening
Cube Roots of Unity
If alpha$\alpha$ is a root of the equation x^2 + x + 1 = 0$x^{2} + x + 1 = 0$ and sum_k=1^nleft(alpha^k + frac1alpha^kright)^2 = 20$\sum_{k=1}^{n}\left(\alpha^{k} + \frac{1}{\alpha^{k}}\right)^{2} = 20$, then n$n$ is equal to
Numerical Answer. Answer: 11 to 11
Solution
### Core Logic
The equation x^2 + x + 1 = 0$x^2 + x + 1 = 0$ has complex roots which are the non-real cube roots of unity. Thus, we can set alpha = omega$\alpha = \omega$ (where omega^3 = 1$\omega^3 = 1$ and 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$).
Let's analyze the general term block T_k = left(omega^k + frac1omega^kright)^2$T_k = \left(\omega^k + \frac{1}{\omega^k}\right)^2$:
T_k = left(omega^k + omega^-kright)^2 = omega^2k + omega^-2k + 2 = omega^2k + omega^k + 2$$T_k = \left(\omega^k + \omega^{-k}\right)^2 = \omega^{2k} + \omega^{-2k} + 2 = \omega^{2k} + \omega^k + 2$$
Because omega^k$\omega^k$ is periodic with period 3$3$, let's examine the values of T_k$T_k$ for different values of k$k$:
- If k$k$ is a multiple of 3$3$ (k=3m$k=3m$): omega^2k = 1, omega^k = 1 implies T_k = 1 + 1 + 2 = 4$\omega^{2k} = 1, \omega^k = 1 \implies T_k = 1 + 1 + 2 = 4$.
- If k$k$ is not a multiple of 3$3$ (k=3m+1$k=3m+1$ or 3m+2$3m+2$): omega^2k + omega^k = -1 implies T_k = -1 + 2 = 1$\omega^{2k} + \omega^k = -1 \implies T_k = -1 + 2 = 1$.
### Step 1: Evaluating periodic blocks
Every block of three consecutive terms (k = 1, 2, 3$k = 1, 2, 3$) contributes exactly:
textSum of a block = 1 + 1 + 4 = 6$$\text{Sum of a block} = 1 + 1 + 4 = 6$$
We want the total summation to equal 20$20$. Let's divide 20$20$ by our block value 6$6$:
20 = 3 times 6 + 2$$20 = 3 \times 6 + 2$$
This means the sum must consist of 3$3$ full periodic blocks plus additional terms that add up to 2$2$.
### Step 2: Determining the final term count n
The number of terms in 3$3$ full blocks is 3 times 3 = 9$3 \times 3 = 9$ terms, giving a sum of 18$18$.
To get the remaining value of 2$2$, we look at the next terms:
- Term 10 (k=10$k=10$, not a multiple of 3) adds 1 implies textTotal = 18 + 1 = 19$1 \implies \text{Total} = 18 + 1 = 19$.
- Term 11 (k=11$k=11$, not a multiple of 3) adds 1 implies textTotal = 19 + 1 = 20$1 \implies \text{Total} = 19 + 1 = 20$.
Hence, the series terminates exactly at n = 11$n = 11$.
### Pattern Recognition
Whenever complex roots of unity or cyclic properties show up inside series sums, group terms into blocks based on the underlying period length (3$3$ here) to convert large sums into simple modular arithmetic arithmetic calculations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers
Class 11 Mathematics: Sequences and Series
Q73
jee_main_2025_04_april_morning
Geometry of Complex Numbers
Let A = \z in mathbbC : |z - 2 - i| = 3\$A = \{z \in \mathbb{C} : |z - 2 - i| = 3\}$, B = \z in mathbbC : operatornameRe(z - iz) = 2\$B = \{z \in \mathbb{C} : \operatorname{Re}(z - iz) = 2\}$ and S = A cap B$S = A \cap B$. Then sum_z in S |z|^2$\sum_{z \in S} |z|^2$ is equal to
Numerical Answer. Answer: 22 to 22
Solution
### Related Formula
Magnitude squared representation:
|z|^2 = x^2 + y^2 quad textfor z = x + iy$$|z|^2 = x^2 + y^2 \quad \text{for } z = x + iy$$
### Core Logic
Convert complex sets into Cartesian forms by setting z = x + iy$z = x + iy$:
Set A$A$: |(x-2) + i(y-1)| = 3 implies (x-2)^2 + (y-1)^2 = 9 quad dots (1)$|(x-2) + i(y-1)| = 3 \implies (x-2)^2 + (y-1)^2 = 9 \quad \dots (1)$
Set B$B$: z - iz = (x+iy) - i(x+iy) = (x+y) + i(y-x)$z - iz = (x+iy) - i(x+iy) = (x+y) + i(y-x)$.
operatornameRe(z - iz) = 2 implies x + y = 2 implies y = 2 - x quad dots (2)$\operatorname{Re}(z - iz) = 2 \implies x + y = 2 \implies y = 2 - x \quad \dots (2)$
### Step 1: Solve System Algebraically
Substitute (2) into (1):
(x - 2)^2 + (2 - x - 1)^2 = 9 implies (x - 2)^2 + (1 - x)^2 = 9$$(x - 2)^2 + (2 - x - 1)^2 = 9 \implies (x - 2)^2 + (1 - x)^2 = 9$$
x^2 - 4x + 4 + 1 - 2x + x^2 = 9 implies 2x^2 - 6x - 4 = 0 implies x^2 - 3x - 2 = 0$$x^2 - 4x + 4 + 1 - 2x + x^2 = 9 \implies 2x^2 - 6x - 4 = 0 \implies x^2 - 3x - 2 = 0$$
Roots are x_1,2 = frac3 pm sqrt172$x_{1,2} = \frac{3 \pm \sqrt{17}}{2}$.
Correspondingly, y = 2 - x implies y_1,2 = frac1 mp sqrt172$y = 2 - x \implies y_{1,2} = \frac{1 \mp \sqrt{17}}{2}$.
### Step 2: Evaluate Sum of Square Magnitudes
Since S$S$ consists of the two intersection points z_1, z_2$z_1, z_2$:
sum_z in S |z|^2 = (x_1^2 + y_1^2) + (x_2^2 + y_2^2) = (x_1^2 + x_2^2) + (y_1^2 + y_2^2)$$\sum_{z \in S} |z|^2 = (x_1^2 + y_1^2) + (x_2^2 + y_2^2) = (x_1^2 + x_2^2) + (y_1^2 + y_2^2)$$
Using identities from quadratic equation x^2 - 3x - 2 = 0$x^2 - 3x - 2 = 0$ (x_1+x_2 = 3, x_1x_2 = -2$x_1+x_2 = 3, x_1x_2 = -2$):
x_1^2 + x_2^2 = (3)^2 - 2(-2) = 13$x_1^2 + x_2^2 = (3)^2 - 2(-2) = 13$.
Since y = 2-x$y = 2-x$, y^2 = 4 - 4x + x^2 implies y_1^2 + y_2^2 = 8 - 4(3) + 13 = 9$y^2 = 4 - 4x + x^2 \implies y_1^2 + y_2^2 = 8 - 4(3) + 13 = 9$.
sum_z in S |z|^2 = 13 + 9 = 22$$\sum_{z \in S} |z|^2 = 13 + 9 = 22$$
### Pattern Recognition
Avoid explicitly using radical root approximations. Summing symmetric expressions directly through standard Vieta coefficient sum shortcuts preserves clean fractions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
Q61
jee_main_2025_07_april_evening
Locus of a Complex Number
If the locus of z in C$z \in C$, such that operatorname R e left(frac z - 12 z + mathrm iright) + operatorname R e left(frac bar z - 12 bar z - mathrm iright) = 2$\operatorname {R e} \left(\frac {z - 1}{2 z + \mathrm {i}}\right) + \operatorname {R e} \left(\frac {\bar {z} - 1}{2 \bar {z} - \mathrm {i}}\right) = 2$, is a circle of radius r$r$ and center (a, b)$(a, b)$ then frac15abr^2$\frac{15ab}{r^2}$ is equal to:
- A. 24$24$
- B. 12$12$
- C. 18$18$
- D. 16$16$
Solution
### Related Formula
For a complex number w$w$, operatornameRe(w) = operatornameRe(barw)$\operatorname{Re}(w) = \operatorname{Re}(\bar{w})$. Hence:
operatornameRe(w) + operatornameRe(barw) = 2operatornameRe(w)$$\operatorname{Re}(w) + \operatorname{Re}(\bar{w}) = 2\operatorname{Re}(w)$$
### Core Logic
Notice that fracbarz - 12barz - i$\frac{\bar{z} - 1}{2\bar{z} - i}$ is the exact complex conjugate of fracz - 12z + i$\frac{z - 1}{2z + i}$.
Thus, the given equation simplifies directly via complex identities to:
2operatornameReleft(fracz - 12z + iright) = 2 implies operatornameReleft(fracz - 12z + iright) = 1$$2\operatorname{Re}\left(\frac{z - 1}{2z + i}\right) = 2 \implies \operatorname{Re}\left(\frac{z - 1}{2z + i}\right) = 1$$
### Step 1: Substitute z = x + iy
Let z = x + iy$z = x + iy$:
frac(x - 1) + iy2x + i(2y + 1)$$\frac{(x - 1) + iy}{2x + i(2y + 1)}$$
To find the real part, multiply numerator and denominator by the conjugate of the denominator:
operatornameReleft[ frac((x - 1) + iy)(2x - i(2y + 1))4x^2 + (2y + 1)^2 right] = 1$$\operatorname{Re}\left[ \frac{((x - 1) + iy)(2x - i(2y + 1))}{4x^2 + (2y + 1)^2} \right] = 1$$
frac2x(x - 1) + y(2y + 1)4x^2 + (2y + 1)^2 = 1$$\frac{2x(x - 1) + y(2y + 1)}{4x^2 + (2y + 1)^2} = 1$$
### Step 2: Expand and Arrange Circle Equation
Expanding the expression:
2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$$2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$$
2x^2 + 2y^2 + 2x + 3y + 1 = 0$$2x^2 + 2y^2 + 2x + 3y + 1 = 0$$
Dividing full equation by 2:
x^2 + y^2 + x + frac32y + frac12 = 0$$x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$$
### Step 3: Extract Center and Radius
textCenter (a, b) = left(-frac12, -frac34right)$$\text{Center } (a, b) = \left(-\frac{1}{2}, -\frac{3}{4}\right)$$
r^2 = g^2 + f^2 - c = left(frac12right)^2 + left(frac34right)^2 - frac12 = frac14 + frac916 - frac12 = frac516$$r^2 = g^2 + f^2 - c = \left(\frac{1}{2}\right)^2 + \left(\frac{3}{4}\right)^2 - \frac{1}{2} = \frac{1}{4} + \frac{9}{16} - \frac{1}{2} = \frac{5}{16}$$
Evaluating frac15abr^2$\frac{15ab}{r^2}$:
frac15 cdot left(-frac12right) cdot left(-frac34right)frac516 = fracfrac458frac516 = 18$$\frac{15 \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{4}\right)}{\frac{5}{16}} = \frac{\frac{45}{8}}{\frac{5}{16}} = 18$$
### Pattern Recognition
Recognizing that operatornameRe(w) + operatornameRe(barw) = 2operatornameRe(w)$\operatorname{Re}(w) + \operatorname{Re}(\bar{w}) = 2\operatorname{Re}(w)$ avoids complex algebraic division on the second fractional expression completely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers
Class 11 Mathematics: Circles
Q56
jee_main_2025_24_jan_morning
Algebraic Properties of Complex Roots
If alpha$\alpha$ and beta$\beta$ are the roots of the equation 2z^2 - 3z - 2i = 0$2z^{2} - 3z - 2i = 0$ , where i = sqrt-1$i = \sqrt{-1}$ , then 16 cdot mathrmReleft(fracalpha^19 + beta^19 + alpha^11 + beta^11alpha^15 + beta^15right) cdot operatornameImleft(fracalpha^19 + beta^19 + alpha^11 + beta^11alpha^15 + beta^15right)$16 \cdot \mathrm{Re}\left(\frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}}\right) \cdot \operatorname{Im}\left(\frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}}\right)$ is equal to :
- A. 398$398$
- B. 312$312$
- C. 409$409$
- D. 441$441$
Solution
### Related Formula
Since alpha$\alpha$ and beta$\beta$ are roots of 2z^2 - 3z - 2i = 0$2z^2 - 3z - 2i = 0$, they satisfy the quadratic equation directly, meaning:
2alpha^2 - 3alpha - 2i = 0 implies 2left(alpha - fracialpharight) = 3 implies alpha - fracialpha = frac32$$2\alpha^2 - 3\alpha - 2i = 0 \implies 2\left(\alpha - \frac{i}{\alpha}\right) = 3 \implies \alpha - \frac{i}{\alpha} = \frac{3}{2}$$
Similarly for beta$\beta$:
beta - fracibeta = frac32$$\beta - \frac{i}{\beta} = \frac{3}{2}$$
### Core Logic
Square the baseline relation to transition to higher exponential powers:
left(alpha - fracialpharight)^2 = left(frac32right)^2 implies alpha^2 - frac1alpha^2 - 2i = frac94$$\left(\alpha - \frac{i}{\alpha}\right)^2 = \left(\frac{3}{2}\right)^2 \implies \alpha^2 - \frac{1}{\alpha^2} - 2i = \frac{9}{4}$$
alpha^2 - frac1alpha^2 = frac94 + 2i$$\alpha^2 - \frac{1}{\alpha^2} = \frac{9}{4} + 2i$$
Squaring once more to isolate the fourth powers:
left(alpha^2 - frac1alpha^2right)^2 = left(frac94 + 2iright)^2$$\left(\alpha^2 - \frac{1}{\alpha^2}\right)^2 = \left(\frac{9}{4} + 2i\right)^2$$
alpha^4 + frac1alpha^4 - 2 = frac8116 - 4 + 9i$$\alpha^4 + \frac{1}{\alpha^4} - 2 = \frac{81}{16} - 4 + 9i$$
alpha^4 + frac1alpha^4 = frac4916 + 9i$$\alpha^4 + \frac{1}{\alpha^4} = \frac{49}{16} + 9i$$
### Step 1: Simplify the Target Expression Fraction
Rearrange the given complex algebraic fraction by factoring out powers:
fracalpha^19 + alpha^11 + beta^19 + beta^11alpha^15 + beta^15 = fracalpha^15left(alpha^4 + frac1alpha^4right) + beta^15left(beta^4 + frac1beta^4right)alpha^15 + beta^15$$\frac{\alpha^{19} + \alpha^{11} + \beta^{19} + \beta^{11}}{\alpha^{15} + \beta^{15}} = \frac{\alpha^{15}\left(\alpha^4 + \frac{1}{\alpha^4}\right) + \beta^{15}\left(\beta^4 + \frac{1}{\beta^4}\right)}{\alpha^{15} + \beta^{15}}$$
Since both alpha$\alpha$ and beta$\beta$ satisfy the exact same symmetric relational identity:
alpha^4 + frac1alpha^4 = beta^4 + frac1beta^4 = frac4916 + 9i$$\alpha^4 + \frac{1}{\alpha^4} = \beta^4 + \frac{1}{\beta^4} = \frac{49}{16} + 9i$$
Substitute this uniform value back into the algebraic expression:
= frac(alpha^15 + beta^15)left(frac4916 + 9iright)alpha^15 + beta^15 = frac4916 + 9i$$= \frac{(\alpha^{15} + \beta^{15})\left(\frac{49}{16} + 9i\right)}{\alpha^{15} + \beta^{15}} = \frac{49}{16} + 9i$$
### Step 2: Extract Real and Imaginary Components
From our simplified expression:
mathrmRe = frac4916$$\mathrm{Re} = \frac{49}{16}$$
operatornameIm = 9$$\operatorname{Im} = 9$$
Now, substitute these into the evaluation formula:
textResult = 16 cdot left(frac4916right) cdot 9 = 49 cdot 9 = 441$$\text{Result} = 16 \cdot \left(\frac{49}{16}\right) \cdot 9 = 49 \cdot 9 = 441$$
### Pattern Recognition
Symmetric rational polynomials in roots alpha, beta$\alpha, \beta$ that can be split into identical numeric multipliers for alpha^n$\alpha^n$ and beta^n$\beta^n$ allow direct cancellation of the polynomial bases without evaluating the individual roots explicitly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers
