Question

Let
\(S = \left\{z∈C : z^2+\overline{z} = 0 \right\}\). Then \(∑_{z∈S}(Re(z)+Im(z))\)
is equal to____.

Answer 1

Correct Answer: 0

To solve the problem, we need to find all complex numbers \( z \) such that \( z^2 + \overline{z} = 0 \).
Let \( z = x + yi \), where \( x, y \in \mathbb{R} \) and \( i \) is the imaginary unit. The conjugate \( \overline{z} = x - yi \).

Substitute into the equation:
\((x + yi)^2 + (x - yi) = 0\)
Expanding \( (x + yi)^2 \):
\( x^2 + 2xyi - y^2 \).
Therefore, the equation becomes:
\( x^2 - y^2 + 2xyi + x - yi = 0 \).
Separate the real and imaginary parts:
\( (x^2 - y^2 + x) + (2xy - y)i = 0 \).
Since the equation must hold for both real and imaginary parts:

1. \( x^2 - y^2 + x = 0 \) (Equation 1)
2. \( 2xy - y = 0 \) (Equation 2)

From Equation 2, factor out \( y \):
\( y(2x - 1) = 0 \).
So, \( y = 0 \) or \( 2x - 1 = 0 \).

If \( y = 0 \), from Equation 1:
\( x^2 + x = 0 \)
\( x(x + 1) = 0 \), leading to \( x = 0 \) or \( x = -1 \).
Thus, possible \( z \) are \( 0 \) and \( -1 \) (as these have \( y = 0 \)).

If \( 2x - 1 = 0 \), then \( x = \frac{1}{2} \). Substituting \( x = \frac{1}{2} \) in Equation 1:
\( \left(\frac{1}{2}\right)^2 - y^2 + \frac{1}{2} = 0 \)
\( \frac{1}{4} - y^2 + \frac{1}{2} = 0 \)
\( \frac{3}{4} = y^2 \)
\( y = \pm\frac{\sqrt{3}}{2} \). Thus, additional solutions are \( \frac{1}{2} + \frac{\sqrt{3}}{2}i \) and \( \frac{1}{2} - \frac{\sqrt{3}}{2}i \).

So \( S = \{0, -1, \frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i\} \). Calculate \( \sum_{z \in S} (Re(z) + Im(z)) \):

- For \( z = 0 \), \( Re(z) + Im(z) = 0 + 0 = 0 \).
- For \( z = -1 \), \( Re(z) + Im(z) = -1 + 0 = -1 \).
- For \( z = \frac{1}{2} + \frac{\sqrt{3}}{2}i \), \( Re(z) + Im(z) = \frac{1}{2} + \frac{\sqrt{3}}{2} \).
- For \( z = \frac{1}{2} - \frac{\sqrt{3}}{2}i \), \( Re(z) + Im(z) = \frac{1}{2} - \frac{\sqrt{3}}{2} \).

Summing these:
\( 0 - 1 + \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} - \frac{\sqrt{3}}{2}\right) = 0 \).

The computed sum is \( 0 \), which fits within the range [0, 0].

Answer 2

The correct answer is 0
\(∵ z^2+\overline{z} = 0\) Let \(z = x+iy\)
\(∴ x^2+y^2+2ixy+x-iy = 0\)
\((x^2-y^2+x)+i(2xy-y) = 0\)
\(∴ x^2+y^2 = 0\) and \((2x-1)y = 0\)
If \(x = +\frac{1}{2}\) then \(y = ±\frac{\sqrt3}{2}\)
And if y = 0 then x = 0, –1
\(∴\) \(z = \{ 0 + 0i, -1 + 0i, \frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i \}\)
\(∴ ∑(R_e(z)+m(z)) = 0\)