We are given the equation:

$(3y - 2)i^{16} + (5 - 2x)i = 0$

Here, $i$ is the imaginary unit, which satisfies $i^2 = -1$. We need to find the real values of $x$ and $y$.

Step 1: Simplify Powers of $i$

First, let's simplify the powers of $i$:

• $i^{16}$: Powers of $i$ cycle every four terms: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. Since $16$ is a multiple of 4, $i^{16} = 1$.

Thus, the equation becomes: $(3y - 2) \cdot 1 + (5 - 2x)i = 0$ Simplifying further: $3y - 2 + (5 - 2x)i = 0$

Step 2: Set Up Real and Imaginary Parts

For this equation to be true, both the real part and the imaginary part must equal zero.

• Real part: $3y - 2 = 0$
• Imaginary part: $5 - 2x = 0$

Step 3: Solve for $x$ and $y$

1. Solving for $y$: $3y - 2 = 0 \quad \Rightarrow \quad 3y = 2 \quad \Rightarrow \quad y = \frac{2}{3}$

2. Solving for $x$: $5 - 2x = 0 \quad \Rightarrow \quad 2x = 5 \quad \Rightarrow \quad x = \frac{5}{2}$

Final Answer

The real values of $x$ and $y$ are: $x = \frac{5}{2}, \quad y = \frac{2}{3}$

Would you like further details, or do you have any questions?

Here are 5 related questions:

1. How does the cyclic nature of $i$ affect solving complex equations?
2. What is the general form of solving equations with real and imaginary parts?
3. How would this equation change if we had $i^8$ instead of $i^{16}$?
4. What are the properties of complex numbers when equated to zero?
5. How can we represent complex numbers geometrically on the complex plane?

Tip: When dealing with powers of $i$, always reduce the exponent mod 4 to simplify calculations.