Let \(\alpha=8-14 \mathrm{i}\), \(\mathrm{A}=\left\{\mathrm{z} \in \mathbb{C}: \frac{\alpha \mathrm{z}-\bar{\alpha} \overline{\mathrm{z}}}{\mathrm{z}^{2}-(\overline{\mathrm{z}})^{2}-112 \mathrm{i}}=1\right\}\) and \(\mathrm{B}=\{\mathrm{z} \in \mathbb{C}:|\mathrm{z}+3 \mathrm{i}|=4\}\). Then \(\sum_{\mathrm{z} \in \mathrm{A} \cap \mathrm{B}}(\operatorname{Re}z-\operatorname{Im} \mathrm{z})\) is equal to ______.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Complex Numbers

Let

α = 8 - 14 i

A = {z \in C : \frac{α z - α ˉ z}{z ^{2} - ( z ) ^{2} - 112 i} = 1}

and

B = {z \in C : ∣ z + 3 i ∣ = 4}

.

Then

\sum_{z \in A \cap B} (Re z - Im z)

is equal to ______.

Answer: 14

Writing

z = x + i y

, a short computation gives

α z - \overset{α}{ˉ} \overset{z}{ˉ} = 2 i (8 y - 14 x)

and

z^{2} - (\overset{z}{ˉ})^{2} = (z - \overset{z}{ˉ}) (z + \overset{z}{ˉ}) = 4 i x y

, so the defining relation becomes

\frac{8 y - 14 x}{4 x y - 112} = 1

, that is

4 x y + 14 x - 8 y - 112 = 0

, which factors as

(x - 4) (y + 7) = 0

. Hence

x = 4

y = - 7

. Intersecting with the circle

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

x = 4

gives

y = - 3

, and

y = - 7

gives

x = 0

. Thus

A \cap B = {4 - 3 i, - 7 i}

, and

\sum (Re z - Im z) = (4 + 3) + (0 + 7) = 14

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer