Question 1181354
Here's the proof of the triangle inequality using the properties you provided:

1. **Start with the square of the absolute value:**

   |z₁ + z₂|² = (z₁ + z₂)(z₁ + z₂)*  (where * denotes the complex conjugate)

2. **Expand the expression:**

   |z₁ + z₂|² = z₁z₁* + z₁z₂* + z₂z₁* + z₂z₂*

3. **Use the property |z|² = z * z*:**

   |z₁ + z₂|² = |z₁|² + z₁z₂* + z₂z₁* + |z₂|²

4. **Rewrite the middle terms using conjugates:** Notice that z₂z₁* is the conjugate of z₁z₂*.  So we can write:

   z₂z₁* = (z₁z₂*)*

5. **Substitute this back into the equation:**

   |z₁ + z₂|² = |z₁|² + z₁z₂* + (z₁z₂*)* + |z₂|²

6. **Use the property Re(z) = (z + z*)/2:** The middle terms can be expressed using the real part:

   z₁z₂* + (z₁z₂*)* = 2 * Re(z₁z₂*)

7. **Substitute this into the equation:**

   |z₁ + z₂|² = |z₁|² + 2 * Re(z₁z₂*) + |z₂|²

8. **Use the property Re(z) ≤ |z|:**

   |z₁ + z₂|² ≤ |z₁|² + 2|z₁z₂*| + |z₂|²

9. **Use the property |ab| = |a||b|:**

   |z₁ + z₂|² ≤ |z₁|² + 2|z₁||z₂*| + |z₂|²

10. **Since |z| = |z*|, we have:**

    |z₁ + z₂|² ≤ |z₁|² + 2|z₁||z₂| + |z₂|²

11. **Recognize the perfect square:** The right side is a perfect square:

    |z₁ + z₂|² ≤ (|z₁| + |z₂|)²

12. **Take the square root of both sides:** Since absolute values are non-negative, we can take the square root without changing the inequality:

    |z₁ + z₂| ≤ |z₁| + |z₂|

This completes the proof of the triangle inequality.