Question

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

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

Answer 1

The correct option is B $\frac{1}{2}(iz+1)+i$

$f\left(z\right)=g\left(z\right)(z-i)(z+i)+az+b;a,bϵC$
$f\left(i\right)=i\Rightarrow ai+b=i...\left(i\right)$
$f(-i)=1+i\Rightarrow a(-i)+b=1+i...\left(ii\right)$

from(i) and (ii),we get $a=\frac{i}{2},b=\frac{1}{2}+i.$

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