Consider the following code. Assume that copy has been imported.w = [1, 2]z = copy.copy(w)z = copy.deepcopy(w)print(z + w)z = w4. (2 points) How many different objects will exist in the computer’s memory after the code has been executed?Assume that no optimizations have been performed and that garbage collection has been executed. WHY?

After the code has been executed, there will be three different objects in the computer's memory. Here's why:

1. The first object is the list w = [1, 2]. This is the original list that was created.

2. The second object is created when z = copy.copy(w) is executed. The copy() function creates a new object that is a shallow copy of w. A shallow copy means that it creates a new object and inserts references into it to the objects found in the original. However, in this case, since the list w only contains immutable objects (integers), the shallow copy is equivalent to a deep copy.

3. The third object is created when z = copy.deepcopy(w) is executed. The deepcopy() function creates a new object and recursively adds copies into it of the objects found in the original. Again, since the list w only contains immutable objects, the deep copy is equivalent to a shallow copy.

4. When z = w is executed, no new object is created. Instead, z now points to the same object as w.

Therefore, in total, there are three different objects in the computer's memory after the code has been executed.