Official solution
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
\[\sum\limits_{K = 0}^\infty {\left( {{1 \over {{n^k}}}} \right)} = \sum\limits_{K = 0}^\infty {\left( {{1 \over {4{n^k}}}} \right)} \]
\[ = {1 \over 1} + {1 \over 4} + {1 \over {16}} + {1 \over {64}} + ......\]
\[ = {1 \over {1 - {1 \over 4}}} = {4 \over 3}\]