Let S=z C: |z-3|1, z(4+3i)+ z(4-3i)24.

Question

Let $S=\{z\in\mathbb C: |z-3|\le1,\ z(4+3i)+\bar z(4-3i)\le24\}$. If $\alpha+i\beta$ is the point in $S$ which is closest to $4i$, then $25(\alpha+\beta)$ is equal to \_\_\_\_

Answer: 80

Accepted Answer

$z(4+3i)+\bar z(4-3i)=2\operatorname{Re}\big(z(4+3i)\big)=2(4x-3y)=8x-6y\le24 \implies 4x-3y\le12$.

The disk $|z-3|\le1$ is centred at $(3,0)$ with radius 1. Notice the line $4x-3y=12$ passes exactly through the centre $(3,0)$: $4(3)-3(0)=12$ --- so this constraint cuts the disk exactly in half.

The point on the full circle $|z-3|=1$ closest to $(0,4)$ lies along the direction from $(3,0)$ toward $(0,4)$: unit vector $\left(-\dfrac35,\dfrac45\right)$, giving the point
$$\left(3-\frac35,\ \frac45\right)=\left(\frac{12}5,\frac45\right).$$

Checking the constraint: $4\left(\dfrac{12}5\right)-3\left(\dfrac45\right)=\dfrac{48-12}5=\dfrac{36}5\le12$  --- so this point lies in $S$ and is indeed the closest point.

$$\alpha=\frac{12}5,\ \beta=\frac45 \implies 25(\alpha+\beta)=25\left(\frac{16}5\right)=80.$$

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