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 $u = \frac{2z + i}{z - ki}$, where $z = x + iy$ and $k > 0$. If the curve represented by $\operatorname{Re}(u) + \operatorname{Im}(u) = 1$ intersects the $y$-axis at the points $P$ and $Q$ such that $PQ = 5$, then the value of $k$ is:

Let $a, b \in \mathbb{R}$ and $a^2+b^2 \neq 0$. Suppose $S = \{z \in \mathbb{C} : z = \frac{1}{a+ibt}, t \in \mathbb{R}, t \neq 0\}$,where $i = \sqrt{-1}$. If $z = x+iy$ and $z \in S$,then $(x, y)$ lies on:

$\sinh(ix)$ is equal to

$\omega$ is a complex cube root of unity and $Z$ is a complex number satisfying $|Z-1| \leq 2$. The possible values of $r$ such that $|Z-1| \leq 2$ and $|\omega Z - 1 - \omega^2| = r$ have no common solution are

Vedclass Test Series

Exam Paper Generator

Online Exam Module