Dividing f(z) by $z-i$, we obtain the remainder $i$ and dividing it by $z+i$, we get the remainder $1+i$, then remainder upon the division of f(z) by $z^2+1$ is

$\textbf{Step -1: Using remainder theorem to get the following equations.}$

$\text{Given, Dividing }f(z) \text{ by } z - i \text{, we get remainder }i$

$\text{Therefore by remainder theorem, }f(i) = i$ $\dots \text{(i)}$

$\text{Given, Dividing }f(z) \text{ by } z + i \text{, we get remainder }1 + i$

$\text{Therefore by remainder theorem, }f(-i) = 1+i$ $\dots \text{(ii)}$

$\text{Let }R(x) \text{ be the remainder when }f(z)\text{is divided by }z^2 + 1$

$\text{degree of R(z) is 1, hence }R(z) = az+b$

$\text{we can simplify }z^2 + 1 = (z -i) (z+i)$

$\therefore f(z) = Q(z)(z-i)(z+i) + R(z)$

$\textbf{Step -2: Solving for a and b.}$

$\text{From (i) and (ii)}$

$f(i) = i = ai + b$

$\Rightarrow b = i - ai$

$f(-i) = 1+i = -ai +b$

$\Rightarrow -ai + i - ai = 1+i$

$\Rightarrow -2ai = 1$

$\Rightarrow a = -\dfrac{1}{2i}$

$= -\dfrac{1}{2i} \times \dfrac{i}{i}$

$= -\dfrac{i}{2(-1)}$

$\therefore a = \dfrac{i}{2}$

$b = i - ai$

$\Rightarrow b = i - \dfrac{i}{2} (i)$

$\Rightarrow b = i - \dfrac{(-1)}{2}$

$\Rightarrow b = i + \dfrac{1}{2}$

$\therefore R(z) = az+b$

$= \dfrac{i}{2}z + i + \dfrac{1}{2}$

$R(z) = \dfrac{1}{2}(iz + 1) + i$

$\textbf{Hence option B is correct}$