JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\mathrm{n}\) denote the number of solutions of the equation \(z^{2}+3 \bar{z}=0\), where \(\mathrm{z}\) is a complex number. Then the value of \(\sum_{k=0}^{\infty} \frac{1}{n^{k}}\) is equal to:
- A \(1\)
- B \(2\)
- C \(\frac{4}{3}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(z^{2}+3 \bar{z}=0\) \(\text { Put } z=x+i y\) \(\Rightarrow x^{2}-y^{2}+2 i x y+3(x-i y)=0\) \(\Rightarrow\left(x^{2}-y^{2}+3 x\right)+i(2 x y-3 y)=0+i 0\) \(\therefore x^{2}-y^{2}+3 x=0 \quad \ldots \ldots(1)\) \(2 x y-3 y=0 \quad \ldots \cdot\) \(x=\frac{3}{2}, y=0\) Put…
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