Question

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

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

Answer 1

Answer: (B)

$\frac{1}{2}(iz+1)+i$

let function be,$f\left(z\right)=g\left(z\right)(z-i)(z+i)+az+b$, $a,b\in C$
Given, $f\left(i\right)=i⟹ai+b=iand,f(-i)=1+i⟹-ai+b=1+i$
From the above two equations we get,
$a=\frac{1}{2}andb=\frac{1}{2}+i$
Hence the remainder is $az+b=\frac{1}{2}z+(\frac{1}{2}+i)z=\frac{1}{2}(iz+1)+i$