Dividing $f\left(z\right)$  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 $f\left(z\right)$  by $z^2+1$  is

$f\left(z\right)=g\left(z\right)(z-i)(z+i)+az+b;a,b\in C$

Given, $f\left(i\right)=i\Rightarrow ai+b=i$     ----     $\left(1\right)$ and  $f(-i)=1+i$

$\Rightarrow a(-i)+b=1+i$    ---  $\left(2\right)$

From $\left(1\right)$  and $\left(2\right)$ , we have $a=\frac{i}{2},b=\frac{1}{2}+i$

Hence, the required remainder is $az+b=(1/2)iz+(1/2)+i.$