QJEEM23Maaa81: JEE Main 2023 Mathematics - Class 11 JEEM23Maaa81 Solution: Let and . Then is equal to ____________.

Let $z = x + iy$

and $\alpha = 8 - 14i$

${{\alpha z - \overline \alpha \,\overline z } \over {{z^2} - {{\overline z }^2} - 112i}} = 1$

$\therefore$ ${{(16y - 28x)} \over {4xy - 112i}} = 1$

$(16y - 28x + 112)i = 4xy$

$\therefore$ $z = - 7i$ or 4

Now, $z = - 7i$ satisfy B

$B:{x^2} + {(y + 3)^2} = 16$

$A \cap B = (0, - 7)$

${\mathop{\rm Re}\nolimits} z - lm\,z = 7$