Let P=zin C:|z+2-3i|≤1$P=\{z\in\mathbb{C}:|z+2-3i|\le1\}$ and Q=zin C:z(1+i)+ z(1-i)≤-8$Q=\{z\in\mathbb{C}:z(1+i)+\overline{z}(1-i)\le-8\}$. Let in P Q, |z-3+2i|$P \cap Q, |z-3+2i|$ be maximum and minimum at z₁$z_{1}$ and z₂$z_{2}$ respectively. If |z₁|²+2|z₂|²=α+β√(2)$|z_{1}|^{2}+2|z_{2}|^{2}=\alpha+\beta\sqrt{2}$, where α, β$\alpha, \beta$ are integers, then α+β$\alpha+\beta$ equals

Related Formula

Since α$\alpha$ is a root of x² - x + 2 = 0$x^2 - x + 2 = 0$, it must satisfy the equation exactly:

This is an essential root reduction property allowing polynomials of high degrees to be collapsed linearly.

Core Logic

We need to evaluate the expression E = α⁶ + α⁴ + β⁴ - 5α²$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$. Use the substitution α² = α - 2$\alpha^2 = \alpha - 2$ to iteratively depress the powers of α$\alpha$.

Substitute α² = α - 2$\alpha^2 = \alpha - 2$ again into the result:

Now, generate α⁶$\alpha^6$ using α⁴$\alpha^4$:

Substitute α² = α - 2$\alpha^2 = \alpha - 2$ into the result again:

Step 1: Simplify the Full Expression

The symmetry of the roots dictates that β⁴$\beta^4$ behaves identically to α⁴$\alpha^4$. Thus:

Substitute all depressed linear forms back into E = α⁶ + α⁴ + β⁴ - 5α²$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$:

Step 2: Apply Sum of Roots

From the original quadratic equation x² - x + 2 = 0$x^2 - x + 2 = 0$, the sum of roots is:

Substitute this back:

(Note: The condition Im(α) gt Im(β)$Im(\alpha) \gt Im(\beta)$ was a distractor since the expression simplified perfectly symmetrically into α + β$\alpha + \beta$ without needing the individual complex values of the roots).

Pattern Recognition

Never compute De Moivre polar forms for high root powers unless the quadratic has roots like ω$\omega$ or i$i$. Always use the characteristic quadratic relation α² = pα + q$\alpha^2 = p\alpha + q$ to rapidly step down degrees until everything is strictly linear.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Related Formula

Core Logic

Let's first find the roots of the lower degree polynomial:

Group the terms:

Thus, the roots are z = -1$z = -1$, and the roots of z² + z + 1 = 0$z^2 + z + 1 = 0$ which are z = ω, ω²$z = \omega, \omega^2$.

Step 1: Check z = -1

Substitute z = -1$z = -1$ into the first equation z¹⁹⁸⁵ + z¹⁰⁰ + 1 = 0$z^{1985} + z^{100} + 1 = 0$:

So, z = -1$z = -1$ is not a common root.

Step 2: Check z = ω and ω²

Substitute z = ω$z = \omega$:

Reduce the powers modulo 3 (since ω³ = 1$\omega^3 = 1$): 1985 = 3 × 661 + 2 ⇒ ω¹⁹⁸⁵ = ω²$1985 = 3 \times 661 + 2 \Rightarrow \omega^{1985} = \omega^2$ 100 = 3 × 33 + 1 ⇒ ω¹⁰⁰ = ω¹ = ω$100 = 3 \times 33 + 1 \Rightarrow \omega^{100} = \omega^1 = \omega$ So, ω¹⁹⁸⁵ + ω¹⁰⁰ + 1 = ω² + ω + 1 = 0$\omega^{1985} + \omega^{100} + 1 = \omega^2 + \omega + 1 = 0$. z = ω$z = \omega$ is a common root.

Substitute z = ω²$z = \omega^2$:

3970 = 3 × 1323 + 1 ⇒ ω³⁹⁷⁰ = ω$3970 = 3 \times 1323 + 1 \Rightarrow \omega^{3970} = \omega$ 200 = 3 × 66 + 2 ⇒ ω²⁰⁰ = ω²$200 = 3 \times 66 + 2 \Rightarrow \omega^{200} = \omega^2$ So, ω³⁹⁷⁰ + ω²⁰⁰ + 1 = ω + ω² + 1 = 0$\omega^{3970} + \omega^{200} + 1 = \omega + \omega^2 + 1 = 0$. z = ω²$z = \omega^2$ is also a common root.

Step 3: Conclusion

There are exactly 2 common roots: ω$\omega$ and ω²$\omega^2$.

Pattern Recognition

Whenever z²+z+1$z^2+z+1$ emerges as a factor, its roots ω$\omega$ and ω²$\omega^2$ can be directly tested in massive degree polynomials using exponent reduction modulo 3.

Chapter Mix

Class 11 Maths: Complex Numbers

Related Formula

|zⁿ| = |z|ⁿ$|z^n| = |z|^n$ | z| = |z|$|\bar{z}| = |z|$

Core Logic

Given the equation: z² = -i z$z^2 = -i\bar{z}$ Taking the modulus on both sides:

Using properties of modulus:

Since |-i| = 1$|-i| = 1$ and | z| = |z|$|\bar{z}| = |z|$: |z|² = |z|$|z|^2 = |z|$ |z|² - |z| = 0$|z|^2 - |z| = 0$

Step 1: Applying the non-zero condition

This gives two possibilities: |z| = 0$|z| = 0$ or |z| = 1$|z| = 1$. Since z = x + iy$z = x + iy$ and xy ≠ 0$xy \neq 0$, neither x$x$ nor y$y$ is zero, which implies |z| ≠ 0$|z| \neq 0$. Therefore, |z| = 1$|z| = 1$. We are asked for |z²|$|z^2|$:

Pattern Recognition

When dealing with equations involving z$z$ and z$\bar{z}$, applying modulus to both sides rapidly simplifies the problem, turning complex equations into simple real algebraic equations in |z|$|z|$.

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Related Formula

Core Logic

For the equation x² - 70x + λ = 0$x^2 - 70x + \lambda = 0$: α + β = 70$\alpha + \beta = 70$ α β = λ$\alpha \beta = \lambda$ Since α, β in N$\alpha, \beta \in \mathbb{N}$, their sum is 70. This gives α(70 - α) = λ$\alpha(70 - \alpha) = \lambda$. We are given λ / 2 N$\lambda / 2 \notin \mathbb{N}$ and λ / 3 N$\lambda / 3 \notin \mathbb{N}$. This means λ$\lambda$ is not divisible by 2 or 3. So λ$\lambda$ must be an odd number and not a multiple of 3. Thus, neither α$\alpha$ nor β$\beta$ can be a multiple of 2 or 3.

Step 1: Finding minimum lambda

We need to find the minimum value of λ$\lambda$. λ = α(70 - α)$\lambda = \alpha(70 - \alpha)$. This parabola opens downwards, so the minimum product occurs when the integers α$\alpha$ and β$\beta$ are as far apart as possible. Let's test small values for α$\alpha$: If α = 1$\alpha = 1$, β = 69 ⇒ λ = 69$\beta = 69 \Rightarrow \lambda = 69$, but 69 = 3 × 23$69 = 3 \times 23$, which is divisible by 3. Rejected. If α = 2$\alpha = 2$, divisible by 2. Rejected. If α = 3$\alpha = 3$, divisible by 3. Rejected. If α = 4$\alpha = 4$, divisible by 2. Rejected. If α = 5$\alpha = 5$, β = 65 ⇒ λ = 325$\beta = 65 \Rightarrow \lambda = 325$. Check divisibility: 325 is odd (not div by 2). 3+2+5 = 10$3+2+5 = 10$ (not div by 3). So minimum λ = 325$\lambda = 325$ with roots α = 5, β = 65$\alpha = 5, \beta = 65$.

Step 2: Evaluating the target expression

We need to compute:

Substitute the values α = 5$\alpha = 5$, β = 65$\beta = 65$, λ = 325$\lambda = 325$:

The result is exactly 60.

Pattern Recognition

Minimizing the product of two numbers with a fixed sum requires them to be as far apart as possible. Divisibility constraints are quickly verified using prime modulus filters (%2 ≠ 0$%2 \neq 0$, %3 ≠ 0$%3 \neq 0$).

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Related Formula

Core Logic

Given z₁+z₂=5$z_1+z_2=5$ and z₁³+z₂³=20+15i$z_1^3+z_2^3=20+15i$.

Now, compute z₁²+z₂²$z_1^2+z_2^2$:

Now, compute z₁⁴+z₂⁴$z_1^4+z_2^4$:

Finally, find magnitude:

Pattern Recognition

Standard algebraic identities recursively applied. Extract sum and product, then ladder up to higher powers.

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations