Let $S=\{z \in C:|z-2| \leq 1, z(1+i)+\bar{z}(1-i) \leq 2\}$. Let $|z-4 i|$ attains minimum and maximum values, respectively, at $\mathrm{z}_{1} \in \mathrm{~S}$ and $\mathrm{z}_{2} \in \mathrm{~S}$. If $5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=\alpha+\beta \sqrt{5}$, where $\alpha$ and $\beta$ are integers, then the value of $\alpha+\beta$ is equal to ____.

Complex Numbers — JEE Main · 2022 · 29 Jun 2022 · Shift 1 Question with Solution

Let

S = {z \in C : ∣ z - 2∣ \leq 1, z (1 + i) + \overset{z}{ˉ} (1 - i) \leq 2}

. Let

∣ z - 4 i ∣

attains minimum and maximum values, respectively, at

z_{1} \in S

and

z_{2} \in S

. If

5 (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}) = α + β 5

, where

α

and

β

are integers, then the value of

α + β

is equal to ____.

Answer: 26

The region

S

is defined by

∣ z - 2∣ \leq 1

(circle centered at

(2, 0)

with radius 1) and

z (1 + i) + \overset{z}{ˉ} (1 - i) \leq 2

.

Letting

z = x + i y

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

, so

x - y \leq 1

.

We want to minimize/maximize

∣ z - 4 i ∣

over this region. The point

P (0, 4)

is outside the region.

The minimum occurs at point

D

on the circle closest to

P

. The line from center

C (2, 0)

P (0, 4)

has slope

- 2

.

The point

D

on the circle in direction of

P

is:

D = (2, 0) + \frac{1}{5} (- 1, 2) = (2 - \frac{1}{5}, \frac{2}{5})

.

So

z_{1} = 2 - \frac{1}{5} + \frac{2 i}{5}

and

∣ z_{1} ∣^{2} = \frac{25 - 4 5}{5}

.

The maximum occurs at

z_{2} = 1

(checking boundary), so

∣ z_{2} ∣^{2} = 1

.

Therefore:

5 (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}) = 25 - 45 + 5 = 30 - 45

.

Thus

α = 30

β = - 4

, and

α + β = 26

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer