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 z² + 1 is

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

Given, f(i) = i

$\Rightarrow \mathrm{ai}+\mathrm{b}=\mathrm{i}$           (1)

and $\mathrm{f}(-\mathrm{i})=1+\mathrm{i}$

$\Rightarrow \mathrm{a}(-\mathrm{i})+\mathrm{b}=1+\mathrm{i}$      (2)

From (l) and (2), we have

$\mathrm{a}=\frac{\mathrm{i}}{2},\mathrm{b}=\frac{1}{2}+\mathrm{i}$

Hence, the required remainder is $az+b=\frac{1}{2}iz+\frac{1}{2}+i$