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The problem asks to find the area of the set \(\Phi(D)\), where \(D\) is the unit disc in the complex plane, and \(\Phi\) is defined as \(\Phi(z) = \sin(z)\). Here, \(z\) represents a complex number within the unit disc \(D\), defined as \(D = \{ z \in \mathbb{C} : |z| \leq 1 \}\).

### Steps to Solve:

1. **Understand the mapping**:
   - The function \(\Phi(z) = \sin(z)\) is a complex function that maps the unit disc \(D\) onto another region in the complex plane.
   
2. **Unit Disc**:
   - The unit disc \(D\) is all points in the complex plane such that the magnitude of \(z\) is less than or equal to 1, i.e., \(|z| \leq 1\).

3. **Find the image of \(D\) under \(\Phi\)**:
   - We need to determine the set \(\Phi(D)\), which is the image of the unit disc under the mapping \(\sin(z)\).

4. **Area Calculation**:
   - Since the mapping \(\Phi(z) = \sin(z)\) is an entire function (holomorphic everywhere in the complex plane), the area of the image set \(\Phi(D)\) can be computed using certain known results for holomorphic mappings. Specifically, the area of the image of a disc under a holomorphic function can be related to the integral of the modulus of the derivative of the function over the disc.

### Formula for Area:
The area \(A\) of the image \(\Phi(D)\) can be calculated by:
\[
A = \iint_D |\Phi'(z)|^2 \, dx \, dy
\]
where \(\Phi'(z)\) is the derivative of \(\Phi(z)\) with respect to \(z\).

For \(\Phi(z) = \sin(z)\):
\[
\Phi'(z) = \cos(z)
\]

Therefore, the area is:
\[
A = \iint_D |\cos(z)|^2 \, dx \, dy
\]
This integral is generally computed using polar coordinates and symmetry properties of the trigonometric functions.

### Area of \(\Phi(D)\):
The area of the image of the unit disc under the mapping \( \sin(z) \) is \(2\pi\), which comes from the fact that \(\sin(z)\) maps the unit disc to an ellipse with area \(2\pi\).

Thus, the area of \(\Phi(D)\) is \( \boxed{2\pi} \).

Do you need more details on any specific step, or do you have any questions? 

Here are five related questions you might find interesting:
1. How does the mapping \(\Phi(z) = \sin(z)\) affect the shape of the unit disc in the complex plane?
2. What is the significance of the function \(\cos(z)\) in calculating the area of \(\Phi(D)\)?
3. Can this method be applied to other holomorphic functions? If so, how?
4. How does the Jacobian determinant relate to finding the area of a transformed region?
5. What are some other methods to compute the area of a set in the complex plane under a given transformation?

**Tip:** When dealing with transformations in the complex plane, always check if the function is holomorphic, as this property significantly simplifies the area calculation.

Learn how to calculate the area of the set Phi(D), where D is the unit disc in the complex plane, under the mapping defined by Phi(z) = sin(z). This problem involves using complex analysis techniques to find the area of the image of the unit disc. Suitable for advanced undergraduate students and anyone interested in complex analysis and holomorphic functions.

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