Find the set of points on the argand plane for which the real part of the complex number `(1 + i)z^2` is positive where `z = x + iy, x , y in R and i = sqrt(-1 )` .

Answer

To solve the problem, we need to find the set of points on the Argand plane (complex plane) for which the real part of the complex number \((1 + i)z^2\) is positive, where \(z = x + iy\) and \(x, y \in \mathbb{R}\).

### Step-by-Step Solution:

1. **Express \(z^2\)**:
\[ z = x + iy \implies z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]

2. **Multiply by \(1 + i\)**:
\[ (1 + i)z^2 = (1 + i)((x^2 - y^2) + (2xy)i) = (x^2 - y^2) + (2xy)i + i(x^2 - y^2) + i^2(2xy) \]
Since \(i^2 = -1\), we have:
\[ (1 + i)z^2 = (x^2 - y^2 - 2xy) + (2xy + x^2 - y^2)i \]

3. **Identify the Real Part**:
The real part of \((1 + i)z^2\) is:
\[ \text{Re}((1 + i)z^2) = x^2 - y^2 - 2xy \]

4. **Set the Condition for Positivity**:
We need:
\[ x^2 - y^2 - 2xy > 0 \]

5. **Rearranging the Inequality**:
Rearranging gives:
\[ x^2 - 2xy - y^2 > 0 \]
This can be factored as:
\[ (x - y)^2 > 0 \]

6. **Analyzing the Inequality**:
The inequality \((x - y)^2 > 0\) implies that \(x - y \neq 0\), or:
\[ x \neq y \]

7. **Finding the Set of Points**:
The set of points where the real part is positive corresponds to the region in the Argand plane excluding the line \(y = x\). Thus, the solution set is:
\[ \{(x, y) \in \mathbb{R}^2 : x \neq y\} \]

### Summary of the Solution:
The set of points on the Argand plane for which the real part of \((1 + i)z^2\) is positive is all points except those on the line \(y = x\).