To find \((1 + i)^2\), let's expand it using the formula for squaring a binomial \((a + b)^2 = a^2 + 2ab + b^2\):

\[
(1 + i)^2 = 1^2 + 2(1)(i) + i^2
\]

Now, let's compute each term:

- \( 1^2 = 1 \)
- \( 2(1)(i) = 2i \)
- \( i^2 = -1 \) (since \( i \) is the imaginary unit, and \( i^2 = -1 \))

Putting it all together:

\[
(1 + i)^2 = 1 + 2i - 1 = 2i
\]

So, \((1 + i)^2 = 2i\).

Would you like any additional details or have any questions?

### Related Questions:
1. What is the cube of \( 1 + i \)?
2. How do you simplify \((2 + 3i)^2\)?
3. What is the magnitude of \( 1 + i \)?
4. How do you multiply two complex numbers?
5. What is the polar form of \( 1 + i \)?

**Tip:** When working with complex numbers, remember that \( i^2 = -1 \), which simplifies many expressions quickly.

Learn how to simplify the square of a complex number (1 + i)² using binomial expansion and the properties of imaginary numbers. This problem explores complex number arithmetic, including i² = -1, and is suitable for students in Grades 11-12.

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