To simplify $(1 + 10i)^2$ into the form $a + bi$, follow these steps:

1. Expand the square: $(1 + 10i)^2 = (1 + 10i)(1 + 10i)$ Use the distributive property: $(1 + 10i)(1 + 10i) = 1 + 10i + 10i + 100i^2$

2. Simplify the terms: Combine the real and imaginary terms: $1 + 10i + 10i + 100i^2 = 1 + 20i + 100i^2$

3. Recall that $i^2 = -1$: Substitute $i^2 = -1$: $1 + 20i + 100(-1) = 1 + 20i - 100$

4. Combine like terms: Simplify the real and imaginary parts: $1 - 100 + 20i = -99 + 20i$

Thus, the simplified form is: $\boxed{-99 + 20i}$

Would you like further steps or a verification? Let me know! Here are some related questions:

1. What is the geometric interpretation of squaring a complex number?
2. How can you use polar form to simplify $(1 + 10i)^2$?
3. What is the modulus and argument of $1 + 10i$?
4. How would the result differ if you used $(1 - 10i)^2$?
5. What happens when you cube $(1 + 10i)$?

Tip: Always remember to apply $i^2 = -1$ when simplifying expressions with imaginary numbers!