JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\alpha=8-14 i , A=\left\{ z \in C : \frac{\alpha z -\bar{\alpha} \overline{ z }}{ z ^2-(\overline{ z })^2-112 i }=1\right\}\) and \(B =\{ z \in C :| z +3 i |=4\}\) Then \(\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)\) is equal to \(...............\).
- A \(14\)
- B \(13\)
- C \(12\)
- D \(11\)
Answer & Solution
Correct Answer
(A) \(14\)
Step-by-step Solution
Detailed explanation
\(\alpha=8-14 i\) \(z=x+i y\) \(a z=(8 x+14 y)+i(-14 x+8 y)\) \(z +\overline{ z }=2 x \quad z -\overline{ z }=2 iy\) 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…
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