Solution to Exercise 2: Prove the Triangle Inequality

Let $z, w \in \mathbb{C}$ (the set of complex numbers).

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(a) Prove that $|z + w|^2 = |z|^2 + |w|^2 + 2\operatorname{Re}(z \overline{w})$:

Step 1: Definition of the modulus squared

For any complex number $z$, $|z|^2 = z \overline{z}$. Using this property, we expand $|z + w|^2$: $|z + w|^2 = (z + w)(\overline{z + w})$

Step 2: Expand $\overline{z + w}$:

Using the property that $\overline{z + w} = \overline{z} + \overline{w}$, we write: $|z + w|^2 = (z + w)(\overline{z} + \overline{w})$

Step 3: Distribute:

Expanding the terms, we get: $|z + w|^2 = z\overline{z} + z\overline{w} + w\overline{z} + w\overline{w}$

Step 4: Group terms:

Using the modulus squared property ($|z|^2 = z\overline{z}$ and $|w|^2 = w\overline{w}$): $|z + w|^2 = |z|^2 + |w|^2 + z\overline{w} + w\overline{z}$

Step 5: Combine the cross terms:

The cross terms $z\overline{w}$ and $w\overline{z}$ are conjugates. Their sum is twice the real part: $z\overline{w} + w\overline{z} = 2\operatorname{Re}(z\overline{w})$

Thus: $|z + w|^2 = |z|^2 + |w|^2 + 2\operatorname{Re}(z\overline{w})$

Final Result: $|z + w|^2 = |z|^2 + |w|^2 + 2\operatorname{Re}(z\overline{w})$

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(b) Prove the Triangle Inequality

The triangle inequality states: $|z + w| \leq |z| + |w|$

Step 1: Use the modulus squared result from part (a):

From part (a), we know: $|z + w|^2 = |z|^2 + |w|^2 + 2\operatorname{Re}(z\overline{w})$

Step 2: Upper bound for the real part:

The real part of any complex number is always less than or equal to its modulus. Thus: $\operatorname{Re}(z\overline{w}) \leq |z\overline{w}|$

Using the modulus product property $|z\overline{w}| = |z||w|$, we get: $\operatorname{Re}(z\overline{w}) \leq |z||w|$

Step 3: Substitute this inequality:

Substitute $\operatorname{Re}(z\overline{w}) \leq |z||w|$ into the expression for $|z + w|^2$: $|z + w|^2 \leq |z|^2 + |w|^2 + 2|z||w|$

Step 4: Recognize the square of a sum:

The right-hand side can be rewritten as: $|z|^2 + |w|^2 + 2|z||w| = (|z| + |w|)^2$

Thus: $|z + w|^2 \leq (|z| + |w|)^2$

Step 5: Take the square root:

Since the modulus is always non-negative, taking the square root of both sides gives: $|z + w| \leq |z| + |w|$

Final Result: $|z + w| \leq |z| + |w|$

This proves the triangle inequality.

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Do you have any questions or would you like further clarifications?

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Related Questions:

1. How can the equality case in the triangle inequality be characterized?
2. What is the geometric interpretation of the triangle inequality in the complex plane?
3. How does the triangle inequality extend to higher dimensions or vectors?
4. Can the triangle inequality be applied to integrals in functional analysis?
5. How does the property $|\operatorname{Re}(z)| \leq |z|$ factor into proving inequalities?

Tip: When proving inequalities involving complex numbers, always explore both algebraic manipulation and geometric intuition.