Let z be those complex numbers which satisfy |z+5|4 and z(1+i)+ z(1-i)-10, i=-1.

Question

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

Answer: 48

Accepted Answer

$|z+5|\le4$: disk centred $(-5,0)$ radius 4.

$z(1+i)+\bar z(1-i)=2(x-y)\ge-10 \implies x-y\ge-5$, i.e. $x-y+5\ge0$.

Intersecting $x+5\ge y$ with the circle $(x+5)^2+y^2=16$ (boundary): substituting $y=x+5$: $(x+5)^2+(x+5)^2=16 \implies (x+5)^2=8 \implies x+5=\pm2\sqrt2$.

Boundary points: $A=(-2\sqrt2-5,-2\sqrt2)$, $B=(2\sqrt2-5,2\sqrt2)$.

Maximum of $|z+1|^2=|z-(-1)|^2$ occurs at the farthest of these from $C=(-1,0)$:
$$AC^2=(-2\sqrt2-5+1)^2+(-2\sqrt2)^2=(2\sqrt2+4)^2+8=32+16\sqrt2+16+8=32+16\sqrt2\cdots$$

Following the source's computation: $AC^2=32+16\sqrt2$, so $\alpha=32,\ \beta=16$.

$$\alpha+\beta=48.$$

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