Question 1181354: Prove that |z1 + z2 | < or = |z1|+|z2|. Use |z|^2 = mod z times z, Re(z) = (z + mod z)/ 2, and Re(z) < or = |z|. The inequality is known as the Triangle Inequality.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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.
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