`((cos theta + i sin theta)^(4))/( (sin theta + i cos theta)^(5))` is equal to

Options

A. ` cos theta -i sin theta`
B. cos `9 theta - i sin 9 theta`
C. `sin theta - icos theta`
D. `sin 9 theta - icos 9 theta`

Answer

To solve the expression \(\frac{(\cos \theta + i \sin \theta)^{4}}{(\sin \theta + i \cos \theta)^{5}}\), we can use Euler's formula and properties of complex numbers.

### Step-by-Step Solution:

1. **Rewrite the Numerator and Denominator Using Euler's Formula**:
We know from Euler's formula that:
\[ \cos \theta + i \sin \theta = e^{i \theta} \]
Therefore, the numerator can be rewritten as:
\[ (\cos \theta + i \sin \theta)^{4} = (e^{i \theta})^{4} = e^{4i \theta} \]

For the denominator, we can rewrite \(\sin \theta + i \cos \theta\):
\[ \sin \theta + i \cos \theta = i(\cos \theta - i \sin \theta) = i e^{-i \theta} \]
Thus, the denominator becomes:
\[ (\sin \theta + i \cos \theta)^{5} = (i e^{-i \theta})^{5} = i^{5} (e^{-i \theta})^{5} = i^{5} e^{-5i \theta} \]

2. **Calculate \(i^{5}\)**:
We know that \(i^{2} = -1\), hence:
\[ i^{4} = 1 \quad \text{and} \quad i^{5} = i^{4} \cdot i = 1 \cdot i = i \]
So, we have:
\[ (\sin \theta + i \cos \theta)^{5} = i e^{-5i \theta} \]

3. **Combine the Results**:
Now we can substitute back into our original expression:
\[ \frac{e^{4i \theta}}{i e^{-5i \theta}} = \frac{e^{4i \theta}}{i} \cdot e^{5i \theta} = \frac{e^{4i \theta} \cdot e^{5i \theta}}{i} = \frac{e^{(4 + 5)i \theta}}{i} = \frac{e^{9i \theta}}{i} \]

4. **Simplify the Expression**:
We can express \(\frac{1}{i}\) as \(-i\):
\[ \frac{e^{9i \theta}}{i} = -i e^{9i \theta} \]

5. **Use Euler's Formula Again**:
Now, substitute \(e^{9i \theta}\) back:
\[ -i e^{9i \theta} = -i (\cos(9\theta) + i \sin(9\theta)) = -i \cos(9\theta) - i^2 \sin(9\theta) \]
Since \(i^2 = -1\), we have:
\[ -i \cos(9\theta) + \sin(9\theta) \]
Thus, the final result can be expressed as:
\[ \sin(9\theta) - i \cos(9\theta) \]

### Final Answer:
\[ \sin(9\theta) - i \cos(9\theta) \]