The value of `(Sigma_(k=1)^(4))(sin.(2pik)/(5)-icos.(2pik)/(5))^(4)` is (where i is iota)

Answer

To solve the problem, we need to evaluate the expression:

\[ \sum_{k=1}^{4} \left( \sin\left(\frac{2\pi k}{5}\right) - i \cos\left(\frac{2\pi k}{5}\right) \right)^{4} \]

### Step-by-Step Solution:

**Step 1: Rewrite the expression.**

We start by rewriting the expression inside the summation:

\[ \sin\left(\frac{2\pi k}{5}\right) - i \cos\left(\frac{2\pi k}{5}\right) = -i\left(\cos\left(\frac{2\pi k}{5}\right) + i \sin\left(\frac{2\pi k}{5}\right)\right) \]

Using Euler's formula, we can express this as:

\[ -i e^{i \frac{2\pi k}{5}} \]

**Hint:** Use Euler's formula \( e^{i\theta} = \cos(\theta) + i \sin(\theta) \) to rewrite trigonometric functions in exponential form.

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**Step 2: Substitute into the summation.**

Now we substitute this back into the summation:

\[ \sum_{k=1}^{4} \left(-i e^{i \frac{2\pi k}{5}}\right)^{4} \]

**Step 3: Simplify the expression.**

Calculating the fourth power:

\[ (-i)^{4} \cdot \left(e^{i \frac{2\pi k}{5}}\right)^{4} = 1 \cdot e^{i \frac{8\pi k}{5}} = e^{i \frac{8\pi k}{5}} \]

So the summation becomes:

\[ \sum_{k=1}^{4} e^{i \frac{8\pi k}{5}} \]

**Hint:** Remember that \( (-i)^4 = 1 \) since \( i^2 = -1 \).

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**Step 4: Evaluate the summation.**

Now we need to evaluate:

\[ \sum_{k=1}^{4} e^{i \frac{8\pi k}{5}} = e^{i \frac{8\pi}{5}} + e^{i \frac{16\pi}{5}} + e^{i \frac{24\pi}{5}} + e^{i \frac{32\pi}{5}} \]

Notice that \( e^{i \frac{8\pi}{5}} \), \( e^{i \frac{16\pi}{5}} \), \( e^{i \frac{24\pi}{5}} \), and \( e^{i \frac{32\pi}{5}} \) are the 5th roots of unity (excluding \( k=0 \)). The sum of all 5th roots of unity is zero.

**Hint:** The roots of unity sum to zero: \( \sum_{k=0}^{n-1} e^{i \frac{2\pi k}{n}} = 0 \).

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**Step 5: Conclude the result.**

Since the sum of the 5th roots of unity is zero, we have:

\[ \sum_{k=1}^{4} e^{i \frac{8\pi k}{5}} = -1 \]

Thus, our original expression simplifies to:

\[ -i \cdot (-1) = i \]

Finally, we need to compute:

\[ (i)^{4} = (i^2)^{2} = (-1)^{2} = 1 \]

### Final Answer:

The value of the expression is:

\[ \boxed{1} \]