If the remainders respectively when `f(z) = z^(3)+z^(2)+2z+1` is divided with `z-i` and `z+i` are `1+3i` and `1+i` then find the remainder when `f(z)` is divided with `z^(2)+1` ?

Answer

To find the remainder when \( f(z) = z^3 + z^2 + 2z + 1 \) is divided by \( z^2 + 1 \), we can use the information given about the remainders when \( f(z) \) is divided by \( z - i \) and \( z + i \).

### Step 1: Identify the roots
The roots of \( z^2 + 1 = 0 \) are \( z = i \) and \( z = -i \).

### Step 2: Apply the Remainder Theorem
According to the Remainder Theorem, when \( f(z) \) is divided by \( z - i \), the remainder is \( f(i) \), and when divided by \( z + i \), the remainder is \( f(-i) \).

We are given:
- \( f(i) = 1 + 3i \)
- \( f(-i) = 1 + i \)

### Step 3: Set up the remainder
Since we are dividing by \( z^2 + 1 \), the remainder \( R(z) \) will be a linear polynomial of the form:
\[ R(z) = az + b \]
where \( a \) and \( b \) are constants to be determined.

### Step 4: Use the values of \( f(i) \) and \( f(-i) \)
Now we can substitute \( z = i \) and \( z = -i \) into the expression for \( R(z) \):

1. For \( z = i \):
\[ R(i) = ai + b = 1 + 3i \]
This gives us the equation:
\[ ai + b = 1 + 3i \quad \text{(1)} \]

2. For \( z = -i \):
\[ R(-i) = -ai + b = 1 + i \]
This gives us the equation:
\[ -ai + b = 1 + i \quad \text{(2)} \]

### Step 5: Solve the system of equations
Now we have a system of two equations:

From equation (1):
\[ b = 1 + 3i - ai \quad \text{(3)} \]

Substituting (3) into equation (2):
\[ -ai + (1 + 3i - ai) = 1 + i \]
Simplifying:
\[ -ai + 1 + 3i - ai = 1 + i \]
Combining like terms:
\[ -2ai + 3i = i \]
This simplifies to:
\[ -2ai = i - 3i \]
\[ -2ai = -2i \]
Dividing both sides by -2:
\[ ai = i \]
Thus, \( a = 1 \).

Now substituting \( a = 1 \) back into equation (3):
\[ b = 1 + 3i - i = 1 + 2i \]

### Step 6: Write the remainder
Now we have:
\[ R(z) = az + b = 1z + (1 + 2i) = z + 1 + 2i \]

### Final Answer
The remainder when \( f(z) \) is divided by \( z^2 + 1 \) is:
\[ \boxed{z + 1 + 2i} \]