Let n denote the number of solutions of the equation z^2+3 z=0, where z is a complex numbe…

Question

Let $n$ denote the number of solutions of the equation $z^2+3\bar z=0$, where $z$ is a complex number. Then the value of $\displaystyle\sum_{k=0}^\infty\frac1{n^k}$ is equal to

Accepted Answer

With $z=x+iy$: $(x^2-y^2+3x)+i(2xy-3y)=0$ gives $y=0$ or $x=\dfrac32$.

If $y=0$: $x(x+3)=0\Rightarrow x=0,-3$ (2 solutions). If $x=\dfrac32$: $y^2=\dfrac{27}4\Rightarrow y=\pm\dfrac{3\sqrt3}2$ (2 more solutions). Total $n=4$.

$$\sum_{k=0}^\infty\frac1{4^k}=\frac1{1-\frac14}=\frac43.$$

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