>>829
To solve this problem, we need to determine the elements of the set
that are also in the set
, which contains all rational numbers. Let's analyze each element in
:
1.
(2−1+i3)2019
:
This is a complex number raised to a high power. Notice that
2−1+i3
is a complex number on the unit circle in the complex plane. Specifically, it represents the cube root of unity,
e2πi/3
. Raising this to the 2019th power will give another root of unity, but not a rational number.
2.
ℑ((21+i)2019)
:
The imaginary part of a complex number raised to a high power. Since
21+i
is on the unit circle, raising it to the 2019th power will rotate it on the unit circle, but its imaginary part will generally not be rational.
3.
:
This is a rational number.
4.
:
This is not a rational number because
is irrational.
5.
sin3π
:
sin3π=23
, which is not rational.
6.
:
While
is a common approximation of
, it is a rational number.
Now, let's count the rational numbers in
:
-
-
There are 2 rational numbers in
.
Therefore,
equals 2.
The answer is
.