Question

Let $n$ denote the number of solutions of the equation $z^2+3\overline{z}=0,$ where $z$ is a complex number. Then the value of $\sum _{k=0}^{\infty }\frac{1}{n^k}$ is equal to

• A. $1$
• B. $2$
• C. $\frac{4}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $\frac{4}{3}$

Let $z=x+iy$
Given:
$z^2+3\overline{z}=0\Rightarrow (x^2-y^2+3x)+i(2xy-3y)=0\Rightarrow x^2-y^2+3x=0\text{and}(2x-3)y=0$
Solving
$(2x-3)y=0\Rightarrow y=0\text{or}x=\frac{3}{2}$
When $y=0$
$x^2-y^2+3x=0\Rightarrow x^2+3x=0\Rightarrow x=0,-3$
When $x=\frac{3}{2}$
$\frac{9}{4}-y^2+\frac{9}{2}=0\Rightarrow y^2=\frac{27}{4}\Rightarrow y=\pm \frac{3\sqrt{3}}{2}$
Therefore, the number of solutions are $4$
Now,
$\sum _{k=0}^{\infty }\frac{1}{n^k}=\sum _{k=0}^{\infty }\frac{1}{4^k}=1+\frac{1}{4}+\frac{1}{16}+\cdots =\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$