To compute $(1 + 3i)^2$, we can use the formula for expanding a complex number raised to a power. Specifically:

$(a + bi)^2 = a^2 + 2abi + (bi)^2$

where $a = 1$ and $b = 3$.

Step-by-Step Solution

1. Square the real part: $a^2 = 1^2 = 1$

2. Square the imaginary part: $(bi)^2 = (3i)^2 = 9i^2 = 9(-1) = -9$

3. Multiply $a$ and $bi$ by 2: $2ab = 2 \cdot 1 \cdot 3i = 6i$

4. Combine all terms: $(1 + 3i)^2 = 1 - 9 + 6i = -8 + 6i$

Final Answer

$(1 + 3i)^2 = -8 + 6i$

Would you like further clarification on any of these steps, or do you have any questions?

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Here are some related questions to further explore this topic:

1. How would you find $(2 + 5i)^2$?
2. What is the modulus of the complex number $-8 + 6i$?
3. How can you represent $(1 + 3i)$ in polar form?
4. How does the complex conjugate affect the square of a complex number?
5. What is the result of $(1 - 3i)^2$?

Tip: When squaring a complex number, remember that $i^2 = -1$, which often changes the sign in the imaginary part calculation.