Solving Complex Number Equation \(Z_1^2 + Z_1Z_2 + Z_2^2 = 0\)

We are given two complex numbers, \(Z_1\) and \(Z_2\), that satisfy the equation:

\( Z_1^2 + Z_1Z_2 + Z_2^2 = 0 \)

Our goal is to find the value of the expression \(\frac{1}{2} + \text{Re}\left(\frac{Z_1}{Z_2}\right)\).

Analyzing the Complex Number Equation

First, let's analyze the given condition \(Z_1^2 + Z_1Z_2 + Z_2^2 = 0\).

We can consider the case where \(Z_2 = 0\). If \(Z_2 = 0\), the equation becomes \(Z_1^2 + 0 + 0 = 0\), which implies \(Z_1 = 0\). In this scenario, the term \(\frac{Z_1}{Z_2}\) is undefined (\(\frac{0}{0}\)). Therefore, we assume \(Z_2 \neq 0\) to proceed.

Deriving the Ratio \(\frac{Z_1}{Z_2}\)

Since we assume \(Z_2 \neq 0\), we can divide the entire equation by \(Z_2^2\):

\( \frac{Z_1^2}{Z_2^2} + \frac{Z_1Z_2}{Z_2^2} + \frac{Z_2^2}{Z_2^2} = \frac{0}{Z_2^2} \)

This simplifies to:

\( \left(\frac{Z_1}{Z_2}\right)^2 + \left(\frac{Z_1}{Z_2}\right) + 1 = 0 \)

Let's substitute \(w = \frac{Z_1}{Z_2}\). The equation becomes a standard quadratic equation in terms of \(w\):

\( w^2 + w + 1 = 0 \)

Solving the Quadratic Equation for \(w\)

We can solve this quadratic equation using the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=1\), and \(c=1\).

\( w = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} \)

\( w = \frac{-1 \pm \sqrt{1 - 4}}{2} \)

\( w = \frac{-1 \pm \sqrt{-3}}{2} \)

Since \(\sqrt{-3} = \sqrt{(-1)(3)} = i\sqrt{3}\), the solutions for \(w\) are:

\( w = \frac{-1 \pm i\sqrt{3}}{2} \)

So, the possible values for \(w = \frac{Z_1}{Z_2}\) are:

• \(w_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\)
• \(w_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\)

These are the complex cubic roots of unity (excluding 1).

Calculating the Real Part of \(\frac{Z_1}{Z_2}\)

The real part of a complex number \(a + bi\) is \(a\). For both possible values of \(w\):

• For \(w_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\), the real part is \(\text{Re}(w_1) = -\frac{1}{2}\).
• For \(w_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\), the real part is \(\text{Re}(w_2) = -\frac{1}{2}\).

In both cases, \(\text{Re}\left(\frac{Z_1}{Z_2}\right) = -\frac{1}{2}\).

Finding the Final Value

Now we need to calculate the value of \(\frac{1}{2} + \text{Re}\left(\frac{Z_1}{Z_2}\right)\).

Substituting the value of the real part we found:

\( \frac{1}{2} + \text{Re}\left(\frac{Z_1}{Z_2}\right) = \frac{1}{2} + \left(-\frac{1}{2}\right) \)

\( \frac{1}{2} - \frac{1}{2} = 0 \)

Conclusion

Therefore, the value of the expression \(\frac{1}{2} + \text{Re}\left(\frac{Z_1}{Z_2}\right)\) is 0.