$z^{2}+3 \bar{z}=0$

$\text { Put } z=x+i y$

$\Rightarrow x^{2}-y^{2}+2 i x y+3(x-i y)=0$

$\Rightarrow\left(x^{2}-y^{2}+3 x\right)+i(2 x y-3 y)=0+i 0$

$\therefore x^{2}-y^{2}+3 x=0 \quad \ldots \ldots(1)$

$2 x y-3 y=0 \quad \ldots \cdot$

$x=\frac{3}{2}, y=0$

Put $x=\frac{3}{2}$ in equation $(1)$

$\frac{9}{4}-y^{2}+\frac{9}{2}=0$

$y^{2}=\frac{27}{4} \Rightarrow y=\pm \frac{3 \sqrt{3}}{2}$

$\therefore(x, y)=\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right),\left(\frac{3}{2}, \frac{-3 \sqrt{3}}{2}\right)$

$\text { Put } y=0 \Rightarrow x^{2}-0+3 x=0$

$\quad x=0,-3$

$\therefore \quad(x, y)=(0,0),(-3,0)$

$\therefore \text { No of solutions }=n=4$

$\sum_{k=0}^{\infty}\left(\frac{1}{n^{k}}\right) \quad=\sum_{k=0}^{\infty}\left(\frac{1}{4^{k}}\right)$

$=\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots \ldots$

$=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$