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.