© All Rights reserved @ LearnWithDash
Step 1: Express the Complex Number in Terms of x and y
Let the complex number be $z = x + i y$. Then its conjugate is $\overline{z} = x - i y$. The given equation is:
$z^2 + 3 \overline{z} = 0$.
Step 2: Substitute z and \overline{z} into the Equation
First, compute $z^2$:
\[
z^2 = (x + i y)^2 = x^2 - y^2 + 2 i x y.
\]
Then substitute $z^2$ and $\overline{z} = x - i y$ into the equation:
\[
x^2 - y^2 + 2ixy + 3(x - i y) = 0.
\]
Step 3: Separate Real and Imaginary Parts
Group real and imaginary parts to set each equal to 0:
\[
\underbrace{(x^2 - y^2 + 3x)}_{\text{Real part}} + \, i \,\underbrace{(2 x y - 3y)}_{\text{Imag part}} = 0 + i \cdot 0.
\]
This yields two equations:
\[
\begin{cases}
x^2 - y^2 + 3x = 0 \quad &(1)\\
2xy - 3y = 0 \quad &(2)
\end{cases}
\]
Step 4: Solve the Imaginary Part Equation
From equation (2), $2xy - 3y = 0$, we can factor out $y$:
\[
y(2x - 3) = 0 \quad \Longrightarrow \quad y = 0 \quad \text{or} \quad x = \frac{3}{2}.
\]
Step 5: Case 1 – y = 0
Substitute $y = 0$ into equation (1):
\[
x^2 - (0)^2 + 3x = 0 \quad \Longrightarrow \quad x^2 + 3x = 0 \quad \Longrightarrow \quad x(x + 3) = 0.
\]
Hence, $x = 0$ or $x = -3$. This gives two solutions for $(x,y)$:
\[
(x,y) = (0,0) \quad \text{and} \quad (x,y) = (-3,0).
\]
Step 6: Case 2 – x = 3/2
Substitute $x = \frac{3}{2}$ into equation (1):
\[
\left(\frac{3}{2}\right)^2 - y^2 + 3 \left(\frac{3}{2}\right) = 0.
\]
Compute each term:
\[
\left(\frac{3}{2}\right)^2 = \frac{9}{4}, \quad 3 \left(\frac{3}{2}\right) = \frac{9}{2}.
\]
So the equation becomes:
\[
\frac{9}{4} - y^2 + \frac{9}{2} = 0 \quad \Longrightarrow \quad -y^2 + \frac{9}{4} + \frac{18}{4} = 0,
\]
\[
-y^2 + \frac{27}{4} = 0 \quad \Longrightarrow \quad y^2 = \frac{27}{4}.
\]
Thus,
\[
y = \pm \frac{3\sqrt{3}}{2}.
\]
This gives two solutions for $(x,y)$:
\[
\left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right) \quad \text{and} \quad \left(\frac{3}{2}, -\frac{3\sqrt{3}}{2}\right).
\]
Step 7: Count the Number of Distinct Solutions
Altogether, there are 4 distinct solutions:
\[
(0,0), \quad (-3,0), \quad \left(\tfrac{3}{2}, \tfrac{3\sqrt{3}}{2}\right), \quad \left(\tfrac{3}{2}, -\tfrac{3\sqrt{3}}{2}\right).
\]
Hence, $n = 4$.
Step 8: Evaluate the Given Infinite Series
The series to be evaluated is:
\[
\sum_{k=0}^{\infty} \frac{1}{n^k}.
\]
With $n = 4$, this becomes:
\[
\sum_{k=0}^{\infty} \left(\frac{1}{4}\right)^k.
\]
This is a geometric series with first term $1$ and common ratio $\frac{1}{4}$. The sum of an infinite geometric series with $|r| < 1$ is given by:
\[
\frac{a}{1-r} \quad \text{where} \quad a = 1 \quad \text{and} \quad r = \frac{1}{4}.
\]
Thus,
\[
\sum_{k=0}^{\infty} \left(\frac{1}{4}\right)^k = \frac{1}{1 - \tfrac{1}{4}} = \frac{1}{\tfrac{3}{4}} = \frac{4}{3}.
\]
Answer
The value of the series is $\frac{4}{3}$.
