Answer :
To determine which expression is equivalent to
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5, \][/tex]
we will apply De Moivre's Theorem, which states that for a complex number in the form [tex]\( r(\cos \theta + i \sin \theta) \)[/tex], its [tex]\( n \)[/tex]-th power is given by
[tex]\[ [r(\cos \theta + i \sin \theta)]^n = r^n \left( \cos(n \theta) + i \sin(n \theta) \right). \][/tex]
Given the expression, we set:
[tex]\[ z = \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \][/tex]
We want to find [tex]\( z^5 \)[/tex]:
[tex]\[ z^5 = \left( \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \right)^5 \][/tex]
According to De Moivre's Theorem:
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos \left( 5 \cdot \frac{\pi}{5} \right) + i \sin \left( 5 \cdot \frac{\pi}{5} \right) \right) \][/tex]
Simplifying the terms inside the cosine and sine functions:
[tex]\[ 5 \cdot \frac{\pi}{5} = \pi \][/tex]
Thus,
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Next, we compute [tex]\( \left( \frac{1}{2} \right)^5 \)[/tex]:
[tex]\[ \left( \frac{1}{2} \right)^5 = \frac{1}{32} \][/tex]
Now, we substitute back into the expression:
[tex]\[ z^5 = \frac{1}{32} \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Since [tex]\( \cos(\pi) = -1 \)[/tex] and [tex]\( \sin(\pi) \approx 0 \)[/tex], we have:
[tex]\[ \cos(\pi) = -1 \][/tex]
[tex]\[ \sin(\pi) = 0 \][/tex]
Thus, the expression simplifies to:
[tex]\[ z^5 = \frac{1}{32} \left( -1 + 0i \right) = -\frac{1}{32} \][/tex]
Therefore, the expression equivalent to
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5 \][/tex]
is:
[tex]\[ \frac{1}{32} [\cos (\pi) + i \sin (\pi)] \][/tex]
So, the answer is:
[tex]\[ \boxed{\frac{1}{32} [\cos (\pi) + i \sin (\pi)]} \][/tex]
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5, \][/tex]
we will apply De Moivre's Theorem, which states that for a complex number in the form [tex]\( r(\cos \theta + i \sin \theta) \)[/tex], its [tex]\( n \)[/tex]-th power is given by
[tex]\[ [r(\cos \theta + i \sin \theta)]^n = r^n \left( \cos(n \theta) + i \sin(n \theta) \right). \][/tex]
Given the expression, we set:
[tex]\[ z = \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \][/tex]
We want to find [tex]\( z^5 \)[/tex]:
[tex]\[ z^5 = \left( \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \right)^5 \][/tex]
According to De Moivre's Theorem:
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos \left( 5 \cdot \frac{\pi}{5} \right) + i \sin \left( 5 \cdot \frac{\pi}{5} \right) \right) \][/tex]
Simplifying the terms inside the cosine and sine functions:
[tex]\[ 5 \cdot \frac{\pi}{5} = \pi \][/tex]
Thus,
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Next, we compute [tex]\( \left( \frac{1}{2} \right)^5 \)[/tex]:
[tex]\[ \left( \frac{1}{2} \right)^5 = \frac{1}{32} \][/tex]
Now, we substitute back into the expression:
[tex]\[ z^5 = \frac{1}{32} \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Since [tex]\( \cos(\pi) = -1 \)[/tex] and [tex]\( \sin(\pi) \approx 0 \)[/tex], we have:
[tex]\[ \cos(\pi) = -1 \][/tex]
[tex]\[ \sin(\pi) = 0 \][/tex]
Thus, the expression simplifies to:
[tex]\[ z^5 = \frac{1}{32} \left( -1 + 0i \right) = -\frac{1}{32} \][/tex]
Therefore, the expression equivalent to
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5 \][/tex]
is:
[tex]\[ \frac{1}{32} [\cos (\pi) + i \sin (\pi)] \][/tex]
So, the answer is:
[tex]\[ \boxed{\frac{1}{32} [\cos (\pi) + i \sin (\pi)]} \][/tex]
