The product of all values of `(cosalpha+isinalpha)^(3//5)` is

Options

A. 1
B. `cosalpha+isinalpha`
C. `cos3alpha+isin3alpha`
D. `cos5alpha+isin5alpha`

Answer

To solve the problem of finding the product of all values of \((\cos \alpha + i \sin \alpha)^{\frac{3}{5}}\), we can follow these steps:

### Step 1: Define the expression
Let:
\[ z = (\cos \alpha + i \sin \alpha)^{\frac{3}{5}} \]

### Step 2: Raise both sides to the power of 5
To eliminate the fractional exponent, we raise both sides to the power of 5:
\[ z^5 = \cos \alpha + i \sin \alpha)^3 \]

### Step 3: Use the formula for powers of complex numbers
Using the formula for powers of complex numbers, we know that:
\[ (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \]
Thus, we can write:
\[ z^5 = \cos(3\alpha) + i \sin(3\alpha) \]

### Step 4: Set up the equation
Now we have:
\[ z^5 - (\cos(3\alpha) + i \sin(3\alpha)) = 0 \]

### Step 5: Identify the roots of the equation
This is a polynomial equation of degree 5, which means it has 5 roots. The product of the roots of a polynomial can be found using Vieta's formulas.

### Step 6: Calculate the product of the roots
For a polynomial of the form:
\[ z^5 + a_4 z^4 + a_3 z^3 + a_2 z^2 + a_1 z + a_0 = 0 \]
the product of the roots is given by:
\[ (-1)^n \frac{a_0}{a_n} \]
where \(n\) is the degree of the polynomial. In our case, \(n = 5\), \(a_0 = -(\cos(3\alpha) + i \sin(3\alpha))\), and \(a_n = 1\).

Thus, the product of the roots is:
\[ \text{Product of roots} = -(-(\cos(3\alpha) + i \sin(3\alpha))) = \cos(3\alpha) + i \sin(3\alpha) \]

### Final Answer
The product of all values of \((\cos \alpha + i \sin \alpha)^{\frac{3}{5}}\) is:
\[ \cos(3\alpha) + i \sin(3\alpha) \]

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