Let n denote the number of solutions of the equation

Solution:

$\mathrm{z}^{2}+3 \overline{\mathrm{z}}=0$

Put $z=x+i y$

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

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

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

$2 x y-3 y=0$ ..............(2)

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

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

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

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

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

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

$x=0,-3$

$\therefore(\mathrm{x}, \mathrm{y})=(0,0),(-3,0)$

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

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

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

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