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

Options

A. `(1)/(2)(z+1)+i`
B. `(1)/(2)(iz +1)+i`
C. `(1)/(2)(iz-1)+i`
D. `(1)/(2)(z+i)+1`

Answer

B

Explanation

`f(z) = g(z)(z-i) (z+i) + az + b,a, b in C`
Given , `f(i) = i`
`rArr ai + b = i`
` and f(-i) = 1 +i`
`rArr a(-i) +b = 1 +i`
From (1) and (2), we have
`a= (i)/(2), b= (1)/(2)+i`
Hence, the required remaider is `az + b = (1//2)iz + (1//2) + i` .