Question

Let \(z\) be those complex numbers which satisfy
\(|z+5|\le4\) and \(z(1+i)+\overline{z}(1-i)\ge-10\), \(i=\sqrt{-1}\).
If the maximum value of \(|z+1|^2\) is \(\alpha+\beta\sqrt2\), then the value of \((\alpha+\beta)\) is ____________.

Hint:

The region is the half-disc of the circle with centre \((-5,0)\), radius 4, cut by the line \(x-y=-5\); the farthest point from \((-1,0)\) is an end of the bounding diameter.

Answer 1

Correct Answer: 48

Let \(z=x+iy\). \(|z+5|\le4\): disc with centre \((-5,0)\), radius 4.
\(z(1+i)+\overline z(1-i)=2\,\mathrm{Re}[(x+iy)(1+i)]=2(x-y)\ge-10\Rightarrow x-y\ge-5\), i.e. \(y\le x+5\). This line passes through the centre \((-5,0)\), so the region is a half-disc.
\(|z+1|^2\) is the squared distance from \(Q(-1,0)\), which lies in the region (\(-1-0=-1\ge-5\)). The farthest point of the full disc, \((-9,0)\), violates \(x-y\ge-5\), so the maximum occurs at an end of the diameter on the line: \((-5\pm2\sqrt2,\ \pm2\sqrt2)\).
For \((-5-2\sqrt2,-2\sqrt2)\): \((-4-2\sqrt2)^2+8=16+16\sqrt2+8+8=32+16\sqrt2\).
For \((-5+2\sqrt2,2\sqrt2)\): \(32-16\sqrt2\) (smaller).
So \(\alpha=32,\beta=16\) and \(\alpha+\beta=48\).
Answer: 48