Question

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

• A. $i+z$
• B. $1+z$
• C. $1-z$
• D. None of the above

Answer 1

The correct option is C $1-z$

When $f\left(z\right)$ is divided by $z-i$, remainder $1-i$ is obtained.
Therefore, by remainder theorem: $f\left(i\right)=1-i$. ...(1)
Also, when $f\left(z\right)$ is divided by $z+i$, remainder $1+i$ is obtained.
Therefore, by remainder theorem: $f(-i)=1+i$. ...(2)
Let $R\left(z\right)$ be the remainder obtained, when $f\left(z\right)$ divided by $(z-i)(z+i)$ and $Q\left(z\right)$ be its quotient.
$\therefore f\left(z\right)=Q\left(z\right)(z-i)(z+i)+R\left(z\right)$
The maximum degree of $R\left(z\right)$ is 1.
Let $R\left(z\right)=az+b$.
From $\left(1\right)$ and $\left(2\right)$:$f\left(i\right)=1-i$ &$f(-i)=1+i$
$\Rightarrow 1-i=ai+b\&1+i=-ai+b\Rightarrow a=-1;b=1$
$\therefore R\left(z\right)=1-z$
Hence, option C is correct.