Let n denote the number of solutions of the equation z² + 3$\overline z$ = 0, where z is a complex number. Then the value of $\sum\limits_{k = 0}^\infty {{1 \over {{n^k}}}}$ is equal to :

z² + 3$\overline z$ = 0<br><br>Put z = x + iy<br><br>$\Rightarrow$ x² $-$ y² + 2ixy + 3(x $-$ iy) = 0<br><br>$\Rightarrow$ (x² $-$ y² + 3x) + i(2xy $-$ 3y) = 0 + i0<br><br>$\therefore$ x² $-$ y² + 3x = 0 ..... (1)<br><br>2xy $-$ 3y = 0 ..... (2)<br><br>x = ${3 \over 2}$, y = 0<br><br>Put x = ${3 \over 2}$ in equation (1)<br><br>${9 \over 4} - {y^2} + {9 \over 2} = 0$<br><br>${y^2} = {{27} \over 4} \Rightarrow y = \pm {{3\sqrt 3 } \over 2}$<br><br>$\therefore$ $$(x,y) = \left( {{3 \over 2},{{3\sqrt 3 } \over 2}} \right),\left( {{3 \over 2},{{ - 3\sqrt 3 } \over 2}} \right)$$<br><br>Put y = 0 $\Rightarrow$ x² $-$ 0 + 3x = 0<br><br>x = 0, $-$3<br><br>$\therefore$ (x, y) = (0, 0), ($-$3, 0)<br><br>$\therefore$ No of solutions = n = 4<br><br>$$\sum\limits_{K = 0}^\infty {\left( {{1 \over {{n^k}}}} \right)} = \sum\limits_{K = 0}^\infty {\left( {{1 \over {4{n^k}}}} \right)} $$<br><br>$= {1 \over 1} + {1 \over 4} + {1 \over {16}} + {1 \over {64}} + ......$<br><br>$= {1 \over {1 - {1 \over 4}}} = {4 \over 3}$