To compute $(1 + i)^{12}$, we can use De Moivre's Theorem. Here’s a step-by-step process to solve it:

Step 1: Express $1 + i$ in polar form.

1. Magnitude of $1 + i$:
   $r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$

2. Argument (angle $\theta$):
   Since $1 + i$ is in the first quadrant, the argument is: $\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$ So, $1 + i$ in polar form is: $1 + i = \sqrt{2} \text{cis} \left(\frac{\pi}{4}\right)$ (where $\text{cis}(\theta)$ stands for $\cos(\theta) + i\sin(\theta)$).

Step 2: Use De Moivre’s Theorem

De Moivre’s Theorem states: $(r \text{cis} \theta)^n = r^n \text{cis} (n \theta)$ So for $(1 + i)^{12}$, we have: $(\sqrt{2} \text{cis} \frac{\pi}{4})^{12} = (\sqrt{2})^{12} \text{cis} \left(12 \cdot \frac{\pi}{4}\right)$

Step 3: Simplify the expression

1. Magnitude: $(\sqrt{2})^{12} = (2^{1/2})^{12} = 2^6 = 64$

2. Argument: $12 \cdot \frac{\pi}{4} = 3\pi$ Since $\text{cis}(3\pi) = \cos(3\pi) + i\sin(3\pi) = -1 + 0i$, we have: $\text{cis}(3\pi) = -1$

Step 4: Final Answer

Thus, $(1 + i)^{12} = 64 \cdot (-1) = -64$

So, $(1 + i)^{12} = -64$.

Would you like more details or have any questions?

Related Questions:

1. What is De Moivre's Theorem and how does it apply to complex numbers?
2. How do you convert a complex number from rectangular to polar form?
3. What are the properties of powers of complex numbers?
4. How can you compute $(1 - i)^{12}$ using a similar method?
5. What happens when you apply De Moivre's theorem for non-integer powers?

Tip: When converting complex numbers to polar form, remember to carefully determine the argument based on the quadrant of the complex number.