Let n denote the number of solutions of the equation z ^ 2 + 3 ¯ z...

Given ${z}^{2} + 3\overset{¯}{z} = 0$

Put $z = x + iy,$ then we know that $\overset{¯}{z} = x - iy,$ hence, we have

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

$⇒ {x}^{2} + {i}^{2}{y}^{2} + 2ixy + 3x - 3iy = 0$

We also, know that ${i}^{2} = - 1,$ hence, we get

$⇒ {x}^{2} - {y}^{2} + 2ixy + 3x - 3iy = 0$

$⇒ \left({x}^{2} - {y}^{2} + 3x\right) + i ( 2xy - 3y ) = 0 + i0$

On comparing the real and imaginary parts, we get

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

And $2xy - 3y = 0 \ldots \left(2\right)$

$⇒ y\left(2x - 3\right) = 0$

$⇒ x = \dfrac{3}{2}, y = 0$

Put $x = \dfrac{3}{2}$ in equation \left(1\right),$we get$\dfrac{9}{4} - {y}^{2} + \dfrac{9}{2} = 0$</p><p>$⇒ {y}^{2} = \dfrac{27}{4}$</p><p>$⇒ y = \pm \dfrac{3\sqrt{3}}{2} .$</p><p>$⇒ \left(x, y\right) = \left(\dfrac{3}{2}, \dfrac{3\sqrt{3}}{2}\right), \left(\dfrac{3}{2}, \dfrac{- 3\sqrt{3}}{2}\right) .$</p><p>Now, put$y = 0,$in the equation \left(1\right),$ we get ${x}^{2} + 3x = 0$

$⇒ x = 0, - 3 .$

$∴ \left(x, y\right) = \left(0, 0\right), \left(- 3, 0\right) .$

$∴$ No of solutions$= n = 4$

Now, $\mathop{∑}_{k = 0}^{\infty}\left(\dfrac{1}{{n}^{k}}\right) = \mathop{∑}_{k = 0}^{\infty}\left(\dfrac{1}{{4}^{k}}\right)$

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

The above progression is a geometric progression, with first term $a = 1$ and common ratio $r = \dfrac{1}{4}$ and the sum of infinite terms of the geometric progression is $\dfrac{a}{1 - r}$

Thus, $\mathop{∑}_{k = 0}^{\infty}\left(\dfrac{1}{{n}^{k}}\right) = \dfrac{1}{\left(1 - \dfrac{1}{4}\right)}$

$= \dfrac{1}{\left(\dfrac{3}{4}\right)} = \dfrac{4}{3} .$