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Disc cut by the line z(1+i) + conj(z)(1-i) = c: greatest value of |z - a|^2 in the region

The same question, asked 2 times: 2021 Paper 17*, 2021 Paper 26. Every wording below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 17

    65
    Let C\mathbb{C} be the set of all complex numbers. Let S1={z∈C:∣z−2∣≤1}S_{1}\mathbf{= \{ z \in}\mathbb{C}:|z - 2| \leq 1\} and S2={z∈C:z(1+i)+z‾(1−i)≥4}S_{2}\mathbf{= \{ z \in}\mathbb{C}:z(1 + i) +\overline{z}(1 - i) \geq 4\}. Then, the maximum value of ∣z−52∣2\left| z-\frac{5}{2} \right|^{2} for z∈S1∩S2z \in S_{1}\cap S_{2} is equal to:

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  2. Question 2

    2021 Paper 26

    89
    Let zz be those complex number which satisfy ∣z+5∣≤4|z + 5| \leq 4 and z(1+i)+zˉ(1−i)≥−10,i=−1z(1 + i) +\bar{z}(1 - i) \geq - 10,i =\sqrt{- 1}. If the maximum value of ∣z+1∣2|z + 1|^{2} is α+β2\alpha + \beta\sqrt{2}, then the value of (α+β)(\alpha + \beta) is