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

To solve $ \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 $ \left(2+3i\right)^2 $
\[
\left(2+3i\right)^2 = (2)^2 + 2(2)(3i) + (3i)^2
= 4 + 12i + 9i^2
\]
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?

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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.

Discover how to solve the complex number expression (2+3i)^2 + (2-3i)^2 using algebraic expansion and properties of imaginary numbers. This step-by-step solution explains key concepts such as i^2 = -1 and simplifies the result. Perfect for students in Grades 10-12.

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