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(C) Given equation: $z^{2}+3 \bar{z}=0$.Let $z=x+iy$,where $x, y \in \mathbb{R}$.Substituting into the equation: $(x+iy)^{2}+3(x-iy)=0$.$x^{2}-y^{2}+2

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$\frac{4}{3}$

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(C) Given equation: $z^{2}+3 \bar{z}=0$.Let $z=x+iy$,where $x, y \in \mathbb{R}$.Substituting into the equation: $(x+iy)^{2}+3(x-iy)=0$.$x^{2}-y^{2}+2ixy+3x-3iy=0$.$(x^{2}-y^{2}+3x) + i(2xy-3y) = 0$.Equating real and imaginary parts to zero:$1) \ 2xy-3y=0 \Rightarrow y(2x-3)=0$.This gives $y=0$ or $x=\frac{3}{2}$.Case $1$: If $y=0$,then $x^{2}+3x=0 \Rightarrow x(x+3)=0$,so $x=0$ or $x=-3$. Solutions: $(0,0)$ and $(-3,0)$.Case $2$: If $x=\frac{3}{2}$,then $(\frac{3}{2})^{2}-y^{2}+3(\frac{3}{2})=0$ $\Rightarrow \frac{9}{4}-y^{2}+\frac{9}{2}=0$ $\Rightarrow y^{2}=\frac{27}{4}$ $\Rightarrow y=\pm \frac{3\sqrt{3}}{2}$. Solutions: $(\frac{3}{2}, \frac{3\sqrt{3}}{2})$ and $(\frac{3}{2}, -\frac{3\sqrt{3}}{2})$.Total number of solutions $n=4$.We need to calculate $\sum_{k=0}^{\infty} \frac{1}{n^{k}} = \sum_{k=0}^{\infty} (\frac{1}{4})^{k}$.This is an infinite geometric series with first term $a=1$ and common ratio $r=\frac{1}{4}$.Sum $= \frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

Let $n$ denote the number of solutions of the equation $z^{2}+3 \bar{z}=0$,where $z$ is a complex number. Then the value of $\sum_{k=0}^{\infty} \frac{1}{n^{k}}$ is equal to:

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