Key Concepts and Formulas

• Complex Number Representation: A complex number $z$ can be represented as $z = x + iy$, where $x$ and $y$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$).
• Complex Conjugate: The complex conjugate of $z = x + iy$ is $\overline{z} = x - iy$.
• Infinite Geometric Series: The sum of an infinite geometric series $a + ar + ar^2 + \dots$ is given by $S = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio, provided that $|r| < 1$.

Step 1: Substitute and Expand

We are given the equation $z^2 + 3\overline{z} = 0$. Our goal is to find the number of solutions, $n$, and then evaluate the given summation. We begin by substituting $z = x + iy$ and $\overline{z} = x - iy$ into the equation: $(x + iy)^2 + 3(x - iy) = 0$ Expanding the square, we get: $(x^2 + 2ixy - y^2) + 3(x - iy) = 0$

Step 2: Separate Real and Imaginary Parts

Now, we group the real and imaginary terms: $(x^2 - y^2 + 3x) + i(2xy - 3y) = 0$ Since the right-hand side is 0, which can be written as $0 + 0i$, we can equate the real and imaginary parts to zero.

Step 3: Form a System of Equations

Equating the real and imaginary parts to zero, we get the following system of equations: $x^2 - y^2 + 3x = 0 \quad (*)$ $2xy - 3y = 0 \quad (**)$

Step 4: Solve for y from the Imaginary Part Equation

We solve the imaginary part equation $(**)$ for $y$: $2xy - 3y = 0$ $y(2x - 3) = 0$ This gives us two cases: $y = 0$ or $2x - 3 = 0$, which implies $x = \frac{3}{2}$.

Step 5: Case 1: y = 0

Substitute $y = 0$ into the real part equation $(*)$: $x^2 - 0^2 + 3x = 0$ $x^2 + 3x = 0$ $x(x + 3) = 0$ This gives us $x = 0$ or $x = -3$. Thus, we have two solutions: $z_1 = 0 + 0i = 0$ and $z_2 = -3 + 0i = -3$.

Step 6: Case 2: x = 3/2

Substitute $x = \frac{3}{2}$ into the real part equation $(*)$: $\left(\frac{3}{2}\right)^2 - y^2 + 3\left(\frac{3}{2}\right) = 0$ $\frac{9}{4} - y^2 + \frac{9}{2} = 0$ $\frac{9}{4} - y^2 + \frac{18}{4} = 0$ $\frac{27}{4} - y^2 = 0$ $y^2 = \frac{27}{4}$ $y = \pm\sqrt{\frac{27}{4}} = \pm\frac{3\sqrt{3}}{2}$ Thus, we have two more solutions: $z_3 = \frac{3}{2} + i\frac{3\sqrt{3}}{2}$ and $z_4 = \frac{3}{2} - i\frac{3\sqrt{3}}{2}$.

Step 7: Determine the Number of Solutions, n

We have found four distinct solutions: $z_1 = 0$, $z_2 = -3$, $z_3 = \frac{3}{2} + i\frac{3\sqrt{3}}{2}$, and $z_4 = \frac{3}{2} - i\frac{3\sqrt{3}}{2}$. Therefore, $n = 4$.

Step 8: Evaluate the Summation

We need to evaluate the sum: $\sum_{k=0}^{\infty} \frac{1}{n^k} = \sum_{k=0}^{\infty} \frac{1}{4^k}$ This is an infinite geometric series with first term $a = 1$ and common ratio $r = \frac{1}{4}$. Since $|r| = \frac{1}{4} < 1$, the series converges. The sum is: $S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

Common Mistakes & Tips

• Losing Solutions: When solving equations like $y(2x-3)=0$, avoid dividing by $y$ to prevent losing the solution $y=0$. Factor instead.
• Checking Convergence: Always verify that $|r| < 1$ before applying the infinite geometric series formula.
• Complex Number Simplification: Remember to separate real and imaginary parts correctly when dealing with complex equations.

Summary

We found the number of solutions to the given complex equation to be $n=4$. Then, we evaluated the infinite geometric series $\sum_{k=0}^{\infty} \frac{1}{4^k}$, which converges to $\frac{4}{3}$.

The final answer is $\boxed{4/3}$, which corresponds to option (B).