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4-1.Complex numbers
hard
माना $\alpha=8-14 i, A=\left\{z \in \mathbb{C}: \frac{\alpha z-\bar{\alpha} \bar{z}}{z^2-(\bar{z})^2-112 i}=1\right\}$ तथा $B=\{z \in \mathbb{C}:|z+3 i|=4\}$ हैं तो $\sum_{\mathrm{z} \in \mathrm{A} \cap \mathrm{B}}(\operatorname{Re} z-\operatorname{Im} z)$ बराबर ___________ है।
A
$14$
B
$13$
C
$12$
D
$11$
(JEE MAIN-2023)
Solution
$\alpha=8-14 i$
$z=x+i y$
$a z=(8 x+14 y)+i(-14 x+8 y)$
$z +\overline{ z }=2 x \quad z -\overline{ z }=2 iy$
Set A: $\frac{2 i(-14 x+8 y)}{i(4 x y-112)}=1$
$(x-4)(y+7)=0$
$x=4 \quad \text { or } \quad y=-7$
Set B: $x^2+(y+3)^2=16$
when $x=4 \quad y=-3$
when $y =-7 \quad x =0$
$\therefore A \cap B=\{4-3 i, 0-7 i\}$
So, $\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)=4-(-3)+(0-(-7))=14$
Standard 11
Mathematics
