Let `z=2+3i` then value of `z^5+(barz)^5` is

Options

A. 240
B. 242
C. 244
D. 246

Answer

To solve the problem, we need to calculate \( z^5 + \overline{z}^5 \) where \( z = 2 + 3i \) and \( \overline{z} = 2 - 3i \).

### Step 1: Write down the expressions for \( z^5 \) and \( \overline{z}^5 \)

Given:
- \( z = 2 + 3i \)
- \( \overline{z} = 2 - 3i \)

We need to find:
\[ z^5 + \overline{z}^5 \]

### Step 2: Use the Binomial Theorem to expand \( z^5 \) and \( \overline{z}^5 \)

Using the Binomial Theorem:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

For \( z^5 \):
\[ (2 + 3i)^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (3i)^k \]

For \( \overline{z}^5 \):
\[ (2 - 3i)^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (-3i)^k \]

### Step 3: Combine the expansions

Now, we can combine the two expansions:
\[ z^5 + \overline{z}^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (3i)^k + \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (-3i)^k \]

### Step 4: Identify which terms will cancel out

Notice that:
- For even \( k \), the terms will add up.
- For odd \( k \), the terms will cancel out.

Thus, we only need to consider the even \( k \) terms:
- \( k = 0, 2, 4 \)

### Step 5: Calculate the contributions from even \( k \)

1. **For \( k = 0 \)**:
\[ \binom{5}{0} (2)^5 = 1 \cdot 32 = 32 \]

2. **For \( k = 2 \)**:
\[ \binom{5}{2} (2)^3 (3i)^2 = 10 \cdot 8 \cdot (-9) = -720 \]

3. **For \( k = 4 \)**:
\[ \binom{5}{4} (2)^1 (3i)^4 = 5 \cdot 2 \cdot 81 = 810 \]

### Step 6: Combine the contributions

Now, we sum these contributions:
\[ z^5 + \overline{z}^5 = 32 - 720 + 810 \]
\[ = 32 + 90 = 122 \]

### Final Result

Thus, the value of \( z^5 + \overline{z}^5 \) is:
\[ \boxed{122} \]