Question

Let $z$ be those complex number which satisfies$|z+5|\le 4$ and $z(1+i)+(1-i)\ge -10,i=\sqrt{-1}$. If the maximum value of $|z+1{|}^2$ is$ɑ+\beta \sqrt{2}$, then the value of $ɑ+\beta$ is

Answer 1

Step 1: Determine all the given data

Given that,

we consider $z=x+y$

$|z+5|\le 4$

Squaring on both the sides

${(x+5)}^2+y^2\le 16...\left(1\right)$

Also,

$z(1+i)+(1-i)\ge -10$

Simplifying the equation

$x–y\ge –5...\left(2\right)$

Step 2: Draw conclusions from the diagram

From (1) and (2), the locus of $z$ is the shaded region in the diagram

$|z+1|$ represents the distance of $z$ from $Q(-1,0).$

Clearly, $p$ is the required position of $z$ when $|z+1|$ is the maximum.

Step 3: Find the value of $ɑ+\beta$

$P\equiv (-5–2\sqrt{2},–2\sqrt{2})$ and $Q\equiv (-1,0)$

Therefore, using the distance formula, i.e.,$\mathrm{distance}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$ we get,

$$\begin{gathered} PQ=\sqrt{(-5-2\sqrt{2}-{(-1))}^2+{(-2\sqrt{2}-0)}^2} \\ PQ=\sqrt{{(-5-2\sqrt{2}+1)}^2+(-2\sqrt{2}{)}^2} \\ PQ=\sqrt{{(-4-2\sqrt{2})}^2+{(-2\sqrt{2})}^2} \\ PQ=\sqrt{24+16\sqrt{2}+8} \\ PQ=\sqrt{32+16\sqrt{2}} \\ P{Q}^2=32+16\sqrt{2} \end{gathered}$$

Since, given that$ɑ+\beta \sqrt{2}$

$ɑ=32$

$\beta =16$

$ɑ+\beta =48$

Therefore, the value of $ɑ+\beta$ is $48$.