Animated Solution for Mathematics - Complex Numbers: 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 ${\sum }_{k=0}^{\infty }\frac{1}{n^k}$ is equal to

Imagine you are standing on the Argand plane, a vast, two-dimensional landscape where every point is not just a coordinate $(x,y)$, but a complex number $z=x+iy$. Today, we are going to solve the equation $z^2+3\bar{z}=0$.

It looks innocent, perhaps even simple, but it is a gateway to understanding how complex algebra dances with real-world geometry.

Our first step is to strip away the abstraction. We define $z=x+iy$, where $x$ and $y$ are real numbers. Consequently, the conjugate $\bar{z}$ becomes $x-iy$.

This substitution is our most powerful tool; it allows us to translate the language of complex numbers into the familiar territory of real algebra. By substituting these into our equation, we get:

Now, let us expand the square. Remember that $i^2=-1$. So, $(x+iy{)}^2$ becomes $x^2-y^2+2ixy$.

When we add the $3(x-iy)$ term, our equation transforms into:

This is where the magic happens. We group the real parts and the imaginary parts separately:

For this complex number to be zero, both the real part and the imaginary part must vanish simultaneously. This gives us a system of two real equations:

1) $x^2-y^2+3x=0$

2) $2xy-3y=0$

The second equation, $2xy-3y=0$, is our golden ticket. We can factor it as $y(2x-3)=0$.

This immediately splits our problem into two distinct cases: either $y=0$ or $x=\frac{3}{2}$.

Case 1: If $y=0$, the first equation becomes $x^2+3x=0$, which factors to $x(x+3)=0$. This gives us two solutions: $x=0$ and $x=-3$. Thus, we have the points $(0,0)$ and $(-3,0)$.

Case 2: If $x=\frac{3}{2}$, we substitute this into the first equation:

Simplifying this, we get $\frac{9}{4}-y^2+\frac{9}{2}=0$, which leads to $y^2=\frac{27}{4}$. Therefore, $y=\pm \frac{3\sqrt{3}}{2}$.

This gives us two more points: $(\frac{3}{2},\frac{3\sqrt{3}}{2})$ and $(\frac{3}{2},-\frac{3\sqrt{3}}{2})$.

We have found exactly $n=4$ solutions. The problem now asks us to evaluate the infinite series ${\sum }_{k=0}^{\infty }\frac{1}{n^k}$, which is ${\sum }_{k=0}^{\infty }\frac{1}{4^k}$.

This is a classic infinite Geometric Progression: $1+\frac{1}{4}+\frac{1}{16}+\dots$.

With a first term $a=1$ and a common ratio $r=\frac{1}{4}$, the sum is given by:

Through this journey, we have seen how a single complex equation can unfold into a beautiful geometric structure, ultimately leading us to a precise, elegant numerical result of $4/3$. Never fear the complexity; embrace the geometry behind it.