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Complex Numbers

Dividivng f...

Question

Dividivng $f (z)$ 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 $f (z)$ by $z^{2} + 1$ is

i + z

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1 + z

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1 - z

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None of the above

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Solution

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

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