Let = 8 - 14 i , A = { z : z - z z 2 - ( z ) 2 - 112 i = 1 } and B = { z : | z + 3 i | = 4 } . Then z A B ( Re z - Im z ) is equal to ________. [2023]

Question

Let $α = 8 - 14 i,$ $A = {z \in ℂ : \frac{α z - \bar{α} \bar{z}}{z^{2} - {(\bar{z})}^{2} - 112 i} = 1}$ and $B = {z \in ℂ : | z + 3 i | = 4} .$ Then $\sum_{z \in A \cap B} (Re z - Im z)$ is equal to ________. [2023]

Smart Achievers · Accepted Answer

(14)

Given, $α = 8 - 14 i$

Let $z = x + i y$

$α z = (8 - 14 i) (x + i y) = (8 x + 14 y) + i (- 14 x + 8 y)$

Now, $z + \bar{z} = 2 x$ and $z - \bar{z} = 2 i y$

For $A,$ $\frac{2 i (- 14 x + 8 y)}{i (4 x y - 112)} = 1$

$\Rightarrow (x - 4) (y + 7) = 0 \Rightarrow x = 4 or y = - 7$

For $B,$ $x^{2} + {(y + 3)}^{2} = 16$

When $x = 4$ , then $y = - 3$ ; when $y = - 7$ , then $x = 0$

$∴ A \cap B = {4 - 3 i, 0 - 7 i}$

So, $\sum_{z \in A \cap B} (Re z - Im z) = 4 - (- 3) + (0 - (- 7)) = 7 + 7 = 14$

Solution

Q.
Let $α = 8 - 14 i,$ $A = {z \in ℂ : \frac{α z - \bar{α} \bar{z}}{z^{2} - {(\bar{z})}^{2} - 112 i} = 1}$ and $B = {z \in ℂ : | z + 3 i | = 4} .$ Then $\sum_{z \in A \cap B} (Re z - Im z)$ is equal to ________. [2023]

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Solution

Q. Let α=8-14i, A={z∈ℂ: αz-α¯z¯z2-(z¯)2-112i=1} and B={z∈ℂ:|z+3i|=4}. Then ∑z∈A∩B(Rez-Imz) is equal to ________. [2023]

Ans. (14) Given, α=8-14i Let z=x+iy αz=(8-14i)(x+iy)=(8x+14y)+i(-14x+8y) Now, z+z¯=2x and z-z¯=2iy For A, 2i(-14x+8y)i(4xy-112)=1 ⇒(x-4)(y+7)=0⇒x=4 or y=-7 For B, x2+(y+3)2=16 When x=4, then y=-3; when y=-7, then x=0 ∴A∩B={4-3i, 0-7i} So, ∑z∈A∩B(Rez-Imz)=4-(-3)+(0-(-7))=7+7=14

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Q.
Let $α = 8 - 14 i,$ $A = {z \in ℂ : \frac{α z - \bar{α} \bar{z}}{z^{2} - {(\bar{z})}^{2} - 112 i} = 1}$ and $B = {z \in ℂ : | z + 3 i | = 4} .$ Then $\sum_{z \in A \cap B} (Re z - Im z)$ is equal to ________. [2023]