Math Problem Statement

find (2+3i)2+(23i)2\left(2+3i\right)^2+\left(2-3i\right)^2

Solution

To solve (2+3i)2+(23i)2\left(2+3i\right)^2 + \left(2-3i\right)^2, we can approach it by calculating each term separately and then adding the results.

Step 1: Expand (2+3i)2\left(2+3i\right)^2

= 4 + 12i + 9i^2$$ Since $$i^2 = -1$$, substitute: $$= 4 + 12i + 9(-1) = 4 + 12i - 9 = -5 + 12i$$ ### Step 2: Expand $$\left(2-3i\right)^2$$ $$\left(2-3i\right)^2 = (2)^2 + 2(2)(-3i) + (-3i)^2 = 4 - 12i + 9i^2$$ Again, since $$i^2 = -1$$, substitute: $$= 4 - 12i + 9(-1) = 4 - 12i - 9 = -5 - 12i$$ ### Step 3: Add the two results Now, add $$\left(2+3i\right)^2$$ and $$\left(2-3i\right)^2$$: $$(-5 + 12i) + (-5 - 12i) = -5 - 5 + 12i - 12i = -10$$ Thus, the result is: $$\boxed{-10}$$ Would you like more details on the steps? --- Here are 5 related questions: 1. What is the general form of a complex number? 2. How do you multiply two complex numbers? 3. What is the square of $$3+4i$$? 4. How does one calculate the magnitude of a complex number? 5. Can you explain the geometric interpretation of complex number multiplication? **Tip**: Remember that $$i^2 = -1$$ is key to simplifying complex number expressions.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Algebraic Expansion

Formulas

(a+bi)^2 = a^2 + 2abi + (bi)^2
i^2 = -1

Theorems

Properties of Complex Numbers
Algebraic Expansion

Suitable Grade Level

Grades 10-12