z² + 3$$\overline z $$ = 0

Put z = x + iy

$$\Rightarrow$$ x² $$-$$ y² + 2ixy + 3(x $$-$$ iy) = 0

$$\Rightarrow$$ (x² $$-$$ y² + 3x) + i(2xy $$-$$ 3y) = 0 + i0

$$\therefore$$ x² $$-$$ y² + 3x = 0 ..... (1)

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

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

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

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

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

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

Put y = 0 $$\Rightarrow$$ x² $$-$$ 0 + 3x = 0

x = 0, $$-$$3

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

$$\therefore$$ No of solutions = n = 4

$$\sum\limits_{K = 0}^\infty {\left( {{1 \over {{n^k}}}} \right)} = \sum\limits_{K = 0}^\infty {\left( {{1 \over {4{n^k}}}} \right)} $$

$$ = {1 \over 1} + {1 \over 4} + {1 \over {16}} + {1 \over {64}} + ......$$

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