Let z be those complex numbers which satisfy $|z+5| \leq 4$ and $z(1+i)+\bar{z}(1-i) \geq-10, i=\sqrt{-1}$. If the maximum value of $|z+1|^{2}$ is $\alpha+\beta \sqrt{2}$, then the value of $(\alpha+\beta)$ is ____

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Complex Numbers

Let z be those complex numbers which satisfy

∣ z + 5∣ \leq 4

and

z (1 + i) + \overset{z}{ˉ} (1 - i) \geq - 10, i = - 1

.
If the maximum value of

∣ z + 1 ∣^{2}

α + β 2

, then the value of

(α + β)

is ____

Answer: 48

With

z = x + i y

z (1 + i) + \overset{z}{ˉ} (1 - i) = 2 x - 2 y

, so the second condition is

x - y \geq - 5

, while the first is the closed disc of centre

(- 5, 0)

and radius

4

.
On the bounding circle

z = - 5 + 4 (cos θ + i sin θ)

∣ z + 1 ∣^{2} = 16 (cos θ - 1)^{2} + 16 sin^{2} θ = 32 (1 - cos θ)

and the linear condition becomes

4 cos θ - 4 sin θ \geq 0

, that is

cos θ \geq sin θ

, which holds for

θ \in [- \frac{3 π}{4}, \frac{π}{4}]

.
The function

∣ z + 1 ∣^{2}

is convex, so over this convex region it is largest at a point of the circular boundary, and

32 (1 - cos θ)

increases as

cos θ

falls. The smallest permitted value is

cos θ = - \frac{1}{2}

θ = - \frac{3 π}{4}

, where

z = - 5 - 22 - 22 i

and

∣ z + 1 ∣^{2} = 32 (1 + \frac{1}{2}) = 32 + 162

Hence

α = 32

β = 16

and

α + β = 48

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer