Let α = 8 - 14i, A = \ z ∈ ℂ: - ¯ α ¯ z z ^ 2 - ( ¯ z ) ^ 2 - 112i...

Given:

$A = \left\{z \in \mathbb{C}: \dfrac{\alphaz - \overset{¯}{\alpha}\overset{¯}{z}}{{z}^{2} - ( \overset{¯}{z}{)}^{2} - 112i} = 1\right\}$

$B = \left\{z \in \mathbb{C}: \left|z + 3i\right| = 4\right\}$

$\alpha = 8 - 14i$

Let

$z = x + iy$

So,

$\alphaz = \left(8 - 14i\right)\left(x + iy\right)$

$⇒ \alphaz = ( 8x + 14y ) + i ( - 14x + 8y )$

Also,

$z + \overset{¯}{z} = 2x and$z - \overset{¯}{z} = 2iy$</p><p>Set A:</p><p>$\dfrac{\alphaz - \overset{¯}{\alpha}\overset{¯}{z}}{{z}^{2} - ( \overset{¯}{z}{)}^{2} - 112i} = 1$</p><p>$⇒ \dfrac{2i\left(- 14x + 8y\right)}{\left(z + \overset{¯}{z}\right)\left(z - \overset{¯}{z}\right) - 112i} = 1$</p><p>$⇒ \dfrac{2i\left(- 14x + 8y\right)}{4ixy - 112i} = 1$</p><p>$⇒ \dfrac{- 14x + 8y}{\left(2xy - 56\right)} = 1$</p><p>$⇒ xy + 7x - 4y - 28 = 0$</p><p>$⇒ \left(x - 4\right)\left(y + 7\right) = 0$</p><p>$⇒ x = 4$or$y = - 7$</p><p>Set$B$:</p><p>\left|z + 3i\right| = 4$

$⇒ \left|x + i\left(y + 3\right)\right| = 4$

$⇒ {x}^{2} + {\left(y + 3\right)}^{2} = 16$

When $x = 4 ⇒ y = - 3$

When $y = - 7 ⇒ x = 0$

$∴ A \cap B = \left\{4 - 3i, 0 - 7i\right\}$

So, $\mathop{∑}_{z \in A \cap B}\left(Rez - Im z\right) = 4 - ( - 3 ) + ( 0 - ( - 7 ) ) = 14$