Let's simplify the expression \( 6i^{33} - 15i^{47} \), where \( i \) is the imaginary unit, defined as \( i = \sqrt{-1} \).

### Step 1: Using powers of \( i \)
The powers of \( i \) follow a cyclical pattern:
\[
i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and then it repeats every four powers.}
\]
So, to simplify \( i^{33} \) and \( i^{47} \), we reduce their exponents modulo 4.

#### Simplifying \( i^{33} \):
\[
33 \mod 4 = 1 \quad \Rightarrow \quad i^{33} = i^1 = i
\]

#### Simplifying \( i^{47} \):
\[
47 \mod 4 = 3 \quad \Rightarrow \quad i^{47} = i^3 = -i
\]

### Step 2: Substitute the simplified powers into the expression
Now that we know \( i^{33} = i \) and \( i^{47} = -i \), we can substitute these into the original expression:
\[
6i^{33} - 15i^{47} = 6i - 15(-i)
\]
Simplifying the right-hand side:
\[
6i + 15i = 21i
\]

### Final Answer:
\[
6i^{33} - 15i^{47} = 21i
\]

Would you like further clarification or details on this?

Here are 5 related questions to deepen your understanding:
1. What is the result of \( i^{52} \)?
2. How do powers of \( i \) repeat, and why is it important?
3. How do you find the modulus of a large exponent by a smaller number?
4. What is the general formula for simplifying \( i^n \) for any integer \( n \)?
5. How would the expression change if the coefficients (6 and -15) were different?

**Tip:** The powers of \( i \) cycle every four terms, so you can always reduce large exponents mod 4 to simplify expressions.

This guide explains how to simplify the expression 6i^33 - 15i^47 using the cyclical powers of the imaginary unit (i). Learn how to reduce large exponents modulo 4 and apply complex number rules. Suitable for students in Grades 10-12.

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