JEE Mains · Maths · STD 11 - 4.1 complex nubers
If the locus of \(z \in \mathrm{C}\), such that
\(\operatorname{Re}\left(\frac{z-1}{2 z+\mathrm{i}}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-\mathrm{i}}\right)=2\)
is a circle of radius \(r\) and center \((a, b)\) then \(\frac{15 a b}{r^2}\) is equal to :
- A 24
- B 12
- C 18
- D 16
Answer & Solution
Correct Answer
(C) 18
Step-by-step Solution
Detailed explanation
\(\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-i}\right)=2 \) \( \text {Here, } \frac{z-1}{2 z+i}=\left(\frac{\overline{\bar{z}-1}}{2 \bar{z}-i}\right)=2 \)…
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