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Complex Numbers question

2023 · 29 Jan · Shift 2 · Q41
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  5. /2023 · 29 Jan · Shift 2 · Q41

Complex Numbers question

2023 · 29 Jan · Shift 2 · Q41

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Let α=8−14i,A={z∈c:αz−α‾z‾z2−(z‾)2−112i=1}\alpha = 8 - 14i,A = \left\{ {z \in c:{{\alpha z - \overline \alpha \overline z } \over {{z^2} - {{\left( {\overline z } \right)}^2} - 112i}}=1} \right\}α=8−14i,A={z∈c:z2−(z)2−112iαz−αz​=1} and B={z∈c:∣z+3i∣=4}B = \left\{ {z \in c:\left| {z + 3i} \right| = 4} \right\}B={z∈c:∣z+3i∣=4}. Then ∑z∈A∩B(Reolimitsz−Imolimitsz)\sum\limits_{z \in A \cap B} {({\mathop{\rm Re} olimits} z - {\mathop{\rm Im} olimits} z)}z∈A∩B∑​(Reolimitsz−Imolimitsz) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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