Animated Solution for Mathematics - Complex Numbers: Let z be those complex number which satisfy $|z+5|\le 4$ and $z(1+i)+\bar{z}(1-i)\ge -10$, $i=\sqrt{-1}$ If the maximum value of $|z+1{|}^2$ is $\alpha +\beta \sqrt{2}$, then the value of $(\alpha +\beta )$ is

Welcome, fellow traveler, to the fascinating world of complex numbers. Today, we are not just solving an equation; we are exploring a landscape. Imagine you are standing on the complex plane, a vast, two-dimensional stage where every point $z$ is a coordinate $(x,y)$.

Our problem provides two constraints that define where our complex number $z$ is allowed to exist. The first constraint, $|z+5|\le 4$, is a classic. In the language of geometry, $|z-z_0|\le R$ represents a solid disk centered at $z_0$ with radius $R$.

Here, our center is $z_0=-5$, which corresponds to the point $(-5,0)$, and our radius is $R=4$. So, our complex number $z$ is trapped within a disk of radius $4$ centered at $(-5,0)$.

Unmasking the Hidden Line

Now, let's look at the second, more intimidating constraint: $z(1+i)+\bar{z}(1-i)\ge -10$. If we let $z=x+iy$, then $\bar{z}=x-iy$. Substituting these into the inequality, we get:

Expanding this, we have $(x-y+i(x+y))+(x-y-i(x+y))\ge -10$. Notice how the imaginary parts, $i(x+y)$ and $-i(x+y)$, beautifully cancel each other out.

We are left with $2(x-y)\ge -10$, which simplifies to $x-y\ge -5$, or $y\le x+5$. This is a simple linear inequality where the boundary is the line $y=x+5$.

If you check the center of our circle, $(-5,0)$, you will see that $0=-5+5$, meaning the line passes exactly through the center of our disk. This divides our disk into two equal halves. The condition $y\le x+5$ tells us we are looking at the region below or on this line, defining our feasible region as a semi-disk.

Maximizing the Distance

Our goal is to maximize $|z+1{|}^2$. Geometrically, $|z-(-1)|$ is the distance between $z$ and the fixed point $Q(-1,0)$. Therefore, $|z+1{|}^2$ is the square of that distance.

Let's check where $Q$ is. Plugging $(-1,0)$ into the circle equation $(x+5{)}^2+y^2=16$, we get $(-1+5{)}^2+0^2=4^2=16$. Point $Q$ lies exactly on the boundary of the circle.

To maximize the distance from a point on the boundary of a disk, we should look at the opposite side of the disk. The furthest points will be the intersection of the circle and the line $y=x+5$. Substituting $y=x+5$ into $(x+5{)}^2+y^2=16$:

This yields $x+5=\pm 2\sqrt{2}$, giving us $x=-5\pm 2\sqrt{2}$. The corresponding $y$ values are $y=x+5=\pm 2\sqrt{2}$. We have two points: $P_1(-5+2\sqrt{2},2\sqrt{2})$ and $P_2(-5-2\sqrt{2},-2\sqrt{2})$. Visually, $P_2$ is the point furthest from $Q(-1,0)$.

Now, let's calculate the square of the distance from $P_2(-5-2\sqrt{2},-2\sqrt{2})$ to $Q(-1,0)$. The distance squared is $(x_2-x_1{)}^2+(y_2-y_1{)}^2$. Substituting our coordinates:

This simplifies to $(-4-2\sqrt{2}{)}^2+(-2\sqrt{2}{)}^2$. Expanding the first term, we get:

Adding the second term, which is $8$, we get $24+16\sqrt{2}+8=32+16\sqrt{2}$. We are given that the maximum value is $\alpha +\beta \sqrt{2}$. Comparing terms, we find $\alpha =32$ and $\beta =16$.

The final sum is $\alpha +\beta =32+16=48$.