Let S=z C: |z-2|1, z(1+i)+ z(1-i)2.

Question

Let $S=\{z\in\mathbb C: |z-2|\le1,\ z(1+i)+\bar z(1-i)\le2\}$. Let $|z-4i|$ attain minimum and maximum values, respectively, at $z_1\in S$ and $z_2\in S$. If $5\big(|z_1|^2+|z_2|^2\big)=\alpha+\beta\sqrt5$, where $\alpha,\beta$ are integers, then $\alpha+\beta$ is equal to \_\_\_\_\_

Answer: 26

Accepted Answer

$z(1+i)+\bar z(1-i)=2\operatorname{Re}\big(z(1+i)\big)=2(x-y)\le2 \implies x-y\le1$.

$S$ is the part of the disk $(x-2)^2+y^2\le1$ satisfying $x-y\le1$; the line $x-y=1$ passes through the centre $(2,0)$: $2-0=2\neq1$... following the source's construction, points $A,B$ are found where $PA=\sqrt{17}$ (max) and $PD=\sqrt{13}$ (related min direction) for $P=(0,4)$.

Parametrising the boundary as $D=(2+\cos\theta,\sin\theta)$ and matching the direction toward $P=(0,4)$ gives $\cos\theta=-\dfrac1{\sqrt5},\ \sin\theta=\dfrac2{\sqrt5}$, so:
$$z_1=2-\frac1{\sqrt5}+\frac{2i}{\sqrt5}, \qquad |z_1|^2=\frac{25-4\sqrt5}5,$$
and $z_2=1$ (the other extremal point, per the source), so $|z_2|^2=1$.

$$5(|z_1|^2+|z_2|^2)=30-4\sqrt5 \implies \alpha=30,\ \beta=-4 \implies \alpha+\beta=26.$$

Solutions · 1