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.