Let alpha = 8 - 14i,A = { z in mathbb{C} : fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $\alpha = 8 - 14i$. Let $z = x + iy$. Then $\bar{z} = x - iy$.The equation for set $A$ is $\frac{\alpha z - \bar{\alpha} \bar{z}}{z^2 - \bar

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) Given $\alpha = 8 - 14i$. Let $z = x + iy$. Then $\bar{z} = x - iy$.The equation for set $A$ is $\frac{\alpha z - \bar{\alpha} \bar{z}}{z^2 - \bar{z}^2 - 112i} = 1$.Numerator: $\alpha z - \bar{\alpha} \bar{z} = (8 - 14i)(x + iy) - (8 + 14i)(x - iy) = (8x + 14y + i(-14x + 8y)) - (8x + 14y + i(14x - 8y)) = 2i(-14x + 8y)$.Denominator: $z^2 - \bar{z}^2 = (z - \bar{z})(z + \bar{z}) = (2iy)(2x) = 4ixy$.So,$\frac{2i(-14x + 8y)}{4ixy - 112i} = 1 \implies \frac{2(-14x + 8y)}{4xy - 112} = 1 \implies -28x + 16y = 4xy - 112$.Rearranging: $4xy + 28x - 16y - 112 = 0 \implies 4x(y + 7) - 16(y + 7) = 0 \implies (4x - 16)(y + 7) = 0$.Thus,$x = 4$ or $y = -7$.For set $B$,$|z + 3i| = 4 \implies x^2 + (y + 3)^2 = 16$.Case $1$: If $x = 4$,then $16 + (y + 3)^2 = 16 \implies y = -3$. So $z_1 = 4 - 3i$.Case $2$: If $y = -7$,then $x^2 + (-7 + 3)^2 = 16 \implies x^2 + 16 = 16 \implies x = 0$. So $z_2 = 0 - 7i$.$A \cap B = \{4 - 3i, -7i\}$.Sum: $(\operatorname{Re}(z_1) - \operatorname{Im}(z_1)) + (\operatorname{Re}(z_2) - \operatorname{Im}(z_2)) = (4 - (-3)) + (0 - (-7)) = 7 + 7 = 14$.

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

Similar Questions

Let $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ and $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$. Then the maximum value of $\{|z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B\}$ is

$\sinh(ix)$ is equal to:

If $z$ is a complex number,then the minimum value of $|z| + |z - 1|$ is

The locus of a point $z$ satisfying $|z|^2 = \operatorname{Re}(z)$ is a circle with centre

For all complex numbers $z_1, z_2$ satisfying $|z_1| = 12$ and $|z_2 - 3 - 4i| = 5$,the minimum value of $|z_1 - z_2|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module