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The problem is asking to simplify \((1 + i)^{20}\) using **De Moivre's Theorem**.

### Steps:

1. **Convert \(1 + i\) into polar form:**
   - First, express \(1 + i\) in polar form as \(r(\cos \theta + i \sin \theta)\).
   - The modulus \(r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}\).
   - The argument \(\theta = \arg(1 + i)\), which is the angle in the complex plane. Since \(1 + i\) lies in the first quadrant, \(\theta = \frac{\pi}{4}\).

   So, \(1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)\).

2. **Apply De Moivre's Theorem:**
   - De Moivre’s Theorem states that \((r(\cos \theta + i \sin \theta))^n = r^n (\cos (n\theta) + i \sin (n\theta))\).
   - For \(n = 20\), we have:
     \[
     (1 + i)^{20} = \left(\sqrt{2} \right)^{20} \left(\cos \left(20 \cdot \frac{\pi}{4}\right) + i \sin \left(20 \cdot \frac{\pi}{4}\right)\right)
     \]
     Simplifying:
     \[
     \left(\sqrt{2}\right)^{20} = 2^{10} = 1024
     \]
     And:
     \[
     20 \cdot \frac{\pi}{4} = 5\pi
     \]

3. **Simplify trigonometric terms:**
   \[
   \cos(5\pi) = \cos(\pi) = -1
   \]
   \[
   \sin(5\pi) = \sin(\pi) = 0
   \]
   Thus, \((1 + i)^{20} = 1024 (-1 + 0i) = -1024\).

### Final Answer:
\[
(1 + i)^{20} = -1024
\]

Would you like further details or clarifications?

### Related Questions:
1. How is the modulus of a complex number calculated?
2. What is the geometric interpretation of De Moivre's theorem?
3. Can De Moivre's theorem be used for non-integer powers?
4. How do you find the argument of a complex number in different quadrants?
5. What happens when applying De Moivre's theorem for negative powers?

**Tip:** Always ensure that the argument (angle) of the complex number is correctly determined based on the quadrant.

Using De Moivre's theorem, simplify (1 + i)^20.

Learn how to simplify the expression (1 + i)^20 using De Moivre's Theorem. This step-by-step solution involves converting the complex number into polar form, applying De Moivre's Theorem, and simplifying the trigonometric expressions.

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