Question

$f\left(z\right)$ when divided by $z-i$ gives remainder $i-1;$ when divided by $z+i$ gives remainder $i+1$. When $f\left(z\right)$ is divided by $z^2+1$, the remainder is

• A. $\frac{i}{2}z+(i-\frac{1}{2})$
• B. $\frac{i}{2}z-(i+\frac{1}{2})$
• C. $iz+i$
• D. $iz+1$

Answer 1

Answer: (C)

$iz+i$

Let $Az+B$ be the remainder when $f\left(z\right)$ is divided by $z^2+1$.

then, we have $f\left(z\right)=(z^2+1)g\left(z\right)+Az+B$

Given: $f\left(z\right)$ when divided by $z-i$ gives remainder $i-1$

$\Rightarrow f\left(i\right)=i\Rightarrow (i^2+1)g\left(i\right)+Ai+B=i-1\Rightarrow Ai+B=i-1$ ...(1)

Also, $f\left(z\right)$ when divided by $z+i$ gives remainder $i+1$

$\Rightarrow f(-i)=i+1\Rightarrow (i^2+1)g(-i)-B=i+1\Rightarrow -Ai+B=i+1$ ...(2)

Solving equation (1) and (2), we get

$A=i$ and $B=i$

$\therefore$ remainder is $iz+i$