Question

Dividing $f\left(z\right)$ by $z-i,$ we get the remainder $i$ and dividing it by $z+i,$ we get the remainder $1+i.$ Find the remainder upon the division of $f\left(z\right)$ by z${}^2+1.$

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

Answer 1

The correct option is A $\frac{1}{2}iz+\frac{1}{2}+i$

Remainder when $f\left(z\right)$ is divided by $(z-i)=f\left(i\right)$

Similarly remainder when f(z) is divided by $(z+i)=f(-i)$

According to question $f\left(i\right)=i$ & $f(-i)=1+i$ ............. (1)

Since $(z^2+1)$ is a quadratic expression, therefore remainder where $f\left(z\right)$ is divided by $z^2+1$ will be in general a linear expression.

Let $g\left(z\right)$ be the quotient and $az+b$ the remainder when $g\left(z\right)$ is divided by $z^2+1$ then

$f\left(z\right)=g\left(z\right).(z^2+1)+az+b$ ........ (2)

$\therefore f\left(i\right)=0+ai+b=i$ [from (1)]

and $f(-i)=-ai+b=1+i$ [from (1)]

After solving we get $a=\frac{i}{2}$ and $b=\frac{1}{2}+i$

Hence required result $=az+b=\frac{1}{2}iz+\frac{1}{2}+i$
Ans: A