Suppose `a, b, c in R and C lt 0`. Let z = a + `(b + ic)^(2015) + (b-ic)^(2015)`, then

Options

A. Re (z) = 0
B. Im (z) = 0
C. `Re (z) gt 0, Im (z) lt 0`
D. `Re (z) lt 0, Im (z) gt 0`

Answer

To solve the problem step by step, we start with the expression for \( z \):

\[ z = a + (b + ic)^{2015} + (b - ic)^{2015} \]

### Step 1: Expand using the Binomial Theorem
We can expand both terms \( (b + ic)^{2015} \) and \( (b - ic)^{2015} \) using the binomial theorem, which states:

\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

For \( (b + ic)^{2015} \):
- Let \( x = b \) and \( y = ic \)
- The expansion is:

\[ (b + ic)^{2015} = \sum_{k=0}^{2015} \binom{2015}{k} b^{2015-k} (ic)^k \]

For \( (b - ic)^{2015} \):
- Let \( x = b \) and \( y = -ic \)
- The expansion is:

\[ (b - ic)^{2015} = \sum_{k=0}^{2015} \binom{2015}{k} b^{2015-k} (-ic)^k \]

### Step 2: Combine the two expansions
Now we combine both expansions:

\[ (b + ic)^{2015} + (b - ic)^{2015} = \sum_{k=0}^{2015} \binom{2015}{k} b^{2015-k} \left( (ic)^k + (-ic)^k \right) \]

### Step 3: Analyze the terms
Notice that \( (ic)^k + (-ic)^k \) will yield:
- \( 0 \) for odd \( k \) (since they cancel each other)
- \( 2(b^{2015-k} (ic)^k) \) for even \( k \)

Thus, we only need to consider the even \( k \) terms:

\[ = 2 \sum_{k \text{ even}} \binom{2015}{k} b^{2015-k} (ic)^k \]

### Step 4: Identify the imaginary part
Since \( c < 0 \), the imaginary parts will contribute to \( z \) as follows:

The imaginary parts will cancel out because for every odd \( k \), the terms will be zero. The only contributions to \( z \) will be from the even \( k \) terms, which will be real.

### Step 5: Conclude about the nature of \( z \)
Since all odd terms cancel out and the remaining terms are all real, we conclude that \( z \) is a purely real number.

### Step 6: Final expression for \( z \)
Thus, we can express \( z \) as:

\[ z = a + \text{(real part from the even terms)} \]

### Conclusion
Since \( z \) is purely real, we cannot determine if it is greater than or less than zero without specific values for \( a \), \( b \), and \( c \). Therefore, the imaginary part of \( z \) is zero.