JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let the minimum value \(v_{0}\) of \(v=|=|^{2}+|z-3|^{2}+|z-60|^{2}\), \(z \in C\) is attained at \(z=z_{0}\). Then \(\left|2 z_{0}^{2}-z_{0}^{3}+3\right|^{2}+v_{0}^{2}\) is equal to.
- A \(1000\)
- B \(1024\)
- C \(1105\)
- D \(1196\)
Answer & Solution
Correct Answer
(A) \(1000\)
Step-by-step Solution
Detailed explanation
\(z_{0} =\left(\frac{0+3+0}{3}, \frac{0+6+0}{3}\right)=(1,2)\) \(v_{0}=|1+2 i|^{2}+|1+2 i-3|^{2}+|1+2 i-6 i|^{2}=30\) \(\text { Then }\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}\) \(=\left|2(1+2 i)^{2}-(1-2 i)^{3}+3\right|^{2}+900\)…
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