JEE MAIN - Mathematics (2023 - 29th January Evening Shift - No. 18)
Let $$\alpha = 8 - 14i,A = \left\{ {z \in c:{{\alpha z - \overline \alpha \overline z } \over {{z^2} - {{\left( {\overline z } \right)}^2} - 112i}}=1} \right\}$$ and $$B = \left\{ {z \in c:\left| {z + 3i} \right| = 4} \right\}$$. Then $$\sum\limits_{z \in A \cap B} {({\mathop{\rm Re}\nolimits} z - {\mathop{\rm Im}\nolimits} z)} $$ is equal to ____________.
Answer
14
Explanation
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
$$B:{x^2} + {(y + 3)^2} = 16$$
$$A \cap B = (0, - 7)$$
$${\mathop{\rm Re}\nolimits} z - lm\,z = 7$$
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