JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let the complex numbers \(\alpha\) and \(\frac{1}{\bar{\alpha}}\) lie on the circles \(\left|z-z_0\right|^2=4\) and \(z-\left.z_0\right|^2=16\) respectively, where \(z_0=1+i\). Then, the value of \(100|\alpha|^2\) is.
- A \(30\)
- B \(20\)
- C \(32\)
- D \(22\)
Answer & Solution
Correct Answer
(B) \(20\)
Step-by-step Solution
Detailed explanation
\( \left|z-z_0\right|^2=4 \) \( \Rightarrow\left(\alpha-z_0\right)\left(\bar{\alpha}-\bar{z}_0\right)=4 \) \( \Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_0-z_0 \bar{\alpha}+\left|z_0\right|^2=4 \)…
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