Let S = {z∈C:|z−2|≤1, z(1+i)+z¯(1−i)≤2}. Let|z−4i| attains minimum and maximum values, respectively, at z1∈S and z2∈S. If 5(|z1|2+|z2|2)=α+β5, where and are integers, then the value of + is equal to……………

Let S = ${z \in C : | z - 2 | \leq 1, z (1 + i) + \bar{z} (1 - i) \leq 2} . L e t | z - 4 i |$ attains minimum and maximum values, respectively, at $z_{1} \in S a n d z_{2} \in S .$ If $5 ({| z_{1} |}^{2} + {| z_{2} |}^{2}) = α + β \sqrt[]{5},$ where and are integers, then the value of + is equal to……………

12 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

7 months ago

Let z = x + iy

$| z - 2 | \leq 1 \Rightarrow | x - 2 + i y | \leq 1$

(x – 2)² + y² $\leq 1$

$z (1 + i) + \bar{z} (1 - i) \leq 2$

$(x + i y) (1 + i) + (x - i y) (1 - i) \leq 2$

$x + i x + i y - y + x - i x - i y - y \leq 2$

x – y $\leq 1$

PD will be least as CP – r = DP, general pts on circle with centre (2, 0) is (2 + r cos, 0 + r sin )

here r = 1, (2 + cos , sin ) now slope of CP is $\frac{4 - 0}{0 - 2} = - 2$

tan = 2

so D point will be $(2 - \frac{1}{\sqrt[]{5}}, \frac{2}{\sqrt[]{5}}) A P$ will be the greatest. A(1, 0)

now $| z_{1}^{2} | + | z_{2}^{2} |$

$= {| 1 + 0 i |}^{2} + {| 2 - \frac{1}{\sqrt[]{5}} + \frac{2 i}{\sqrt[]{5}} |}^{2}$

$= 1 + {(2 - \frac{1}{\sqrt[]{5}})}^{2} + \frac{4}{5}$

$= 6 - \frac{4}{\sqrt[]{5}}$

now $5 ({| z_{1} |}^{2} + {| z_{2} |}^{2}) = α + β \sqrt[]{5}$

$5 (6 - \frac{4}{\sqrt[]{5}}) = α + β \sqrt[]{5}$

= 30, = 4

+ = 26

Upvote

Similar Questions for you

If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........

Raj Pandey

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a