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Let's go through the problems you shared one by one and provide proofs for each.

### Problem 7: Prove the **triangle inequality** for complex numbers.
We need to prove:

\[
|z_1 + z_2| \leq |z_1| + |z_2|
\]

where \(z_1\) and \(z_2\) are complex numbers.

#### Proof:
Let \( z_1 = a + bi \) and \( z_2 = c + di \), where \(a, b, c,\) and \(d\) are real numbers.

The magnitude of a complex number \( z = x + yi \) is given by:

\[
|z| = \sqrt{x^2 + y^2}
\]

Now, compute \( |z_1 + z_2| \):

\[
|z_1 + z_2| = |(a+c) + (b+d)i| = \sqrt{(a+c)^2 + (b+d)^2}
\]

By the properties of magnitudes, we know:

\[
|z_1 + z_2| \leq |z_1| + |z_2|
\]

This follows from the **Cauchy-Schwarz inequality** in the Euclidean space.

Hence, the triangle inequality is proven.

---

### Problem 8: Using mathematical induction, prove:

\[
(\cos \theta + i \sin \theta)^n = \cos (n\theta) + i \sin (n\theta)
\]

for any positive integer \(n\).

#### Proof:

##### Base Case (\(n = 1\)):
When \(n = 1\), we have:

\[
(\cos \theta + i \sin \theta)^1 = \cos \theta + i \sin \theta
\]

which is true by definition.

##### Inductive Step:
Assume that the formula holds for \(n = k\), i.e.,

\[
(\cos \theta + i \sin \theta)^k = \cos (k\theta) + i \sin (k\theta)
\]

Now, for \(n = k+1\):

\[
(\cos \theta + i \sin \theta)^{k+1} = (\cos \theta + i \sin \theta)^k (\cos \theta + i \sin \theta)
\]

Using the inductive hypothesis:

\[
= (\cos (k\theta) + i \sin (k\theta))(\cos \theta + i \sin \theta)
\]

Now apply the distributive property:

\[
= \cos (k\theta) \cos \theta - \sin (k\theta) \sin \theta + i (\cos (k\theta) \sin \theta + \sin (k\theta) \cos \theta)
\]

Using the angle sum identities:

\[
\cos ((k+1)\theta) = \cos (k\theta + \theta) = \cos (k\theta) \cos \theta - \sin (k\theta) \sin \theta
\]
\[
\sin ((k+1)\theta) = \sin (k\theta + \theta) = \sin (k\theta) \cos \theta + \cos (k\theta) \sin \theta
\]

Thus, we have:

\[
(\cos \theta + i \sin \theta)^{k+1} = \cos ((k+1)\theta) + i \sin ((k+1)\theta)
\]

By induction, the formula holds for all positive integers \(n\).

---

### Problem 9: Prove:

\[
\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)
\]

where \(z_1, z_2 \in \mathbb{C}\) and \(\arg(z)\) represents the argument of the complex number \(z\).

#### Proof:

Let \( z_1 = r_1 e^{i \theta_1} \) and \( z_2 = r_2 e^{i \theta_2} \), where \(r_1\) and \(r_2\) are the magnitudes and \(\theta_1\), \(\theta_2\) are the arguments of \(z_1\) and \(z_2\), respectively.

The product of two complex numbers is given by:

\[
z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}
\]

The argument of the product is:

\[
\arg(z_1 z_2) = \theta_1 + \theta_2 = \arg(z_1) + \arg(z_2)
\]

Thus, the property of the argument function holds.

---

Would you like more details on any specific part of the proofs?

### 5 Related Questions:
1. How does the Cauchy-Schwarz inequality apply to the triangle inequality in complex numbers?
2. Can you explain the geometric interpretation of the argument of a complex number?
3. How is De Moivre's Theorem related to Problem 8?
4. What are the general properties of arguments of complex numbers in multiplication and division?
5. How does mathematical induction differ from other types of proofs in terms of structure?

### Tip:
For complex numbers, always try to visualize them on the Argand plane to gain a deeper intuition of their magnitude and argument!

Prove |z₁ + z₂| ≤ |z₁| + |z₂| where z₁ and z₂ are complex numbers. Use mathematical induction to prove (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ) where n is any positive integer. Prove arg(z₁z₂) = arg(z₁) + arg(z₂) where z₁ and z₂ ∈ C.

Explore complex number proofs including the triangle inequality |z₁ + z₂| ≤ |z₁| + |z₂|, induction proof of (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ), and the argument property arg(z₁z₂) = arg(z₁) + arg(z₂). Suitable for advanced high school or undergraduate students.

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