Question

Let $S = z ∈ C: |z-3| <= 1$ and $z (4+3i)+z(4-3)≤24.$
If α + iβ is the point in S which is closest to 4i, then 25(α + β) is equal to ______.

Answer 1

The correct answer is 80
Here |z – 3| < 1
$⇒ (x – 3)^2 + y^2 < 1$
And
$z= (4+3i)+\overline{z}(4-3i)$ $≤ 24$
$⇒ 4x-3y ≤ 12$
$tanθ = \frac{4}{3}$

Figure.

∴ Coordinate of P = (3 – cosθ, sinθ)
$=(3-\frac{3}{5},\frac{4}{5})$
$∴ α+iβ = \frac{12}{5}+\frac{4}{5}i$
$∴ 25(α+β)=80$