Let a -Im1- z 2-2 iz - where z is any non-zero complex number- The set A - a - z -1- z - 1 is equal toA- -1-1B- -1-1-C- -0-1 D- -1-0-

The correct option is A (– 1, 1)Given |z|=1 and z ≠± 1 ∴ z = cos θ +isin θ, θ≠ 0, π Now ( 1+z^2/2iz )=1+(cos θ+i sin θ)^2/2i(cos θ+i sin θ )=(1+cos 2θ+i sin 2 ...

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Standard XII

Mathematics

Domain

Let a =Im1+ z...

Question

Let a =Im $(\frac{1 + z^{2}}{2 i z})$ , where z is any non-zero complex number. The set $A = {a : | z | = 1, z \neq \pm 1}$ is equal to

(– 1, 1)

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[–1, 1]

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[0, 1)

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(–1, 0]

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Solution

The correct option is A (– 1, 1)
Given |z|=1 and $z \neq \pm 1$
$∴ z = c o s θ + i s i n θ, θ \neq 0, π$
Now $(\frac{1 + z^{2}}{2 i z}) = \frac{1 + (c o s θ + i s i n θ)^{2}}{2 i (c o s θ + i s i n θ)} = \frac{(1 + c o s 2 θ + i s i n 2 θ) (- s i n θ - i c o s θ)}{2}$
$∴ a = I m (\frac{1 + z^{2}}{2 i z}) = \frac{- 1}{2} [c o s θ (1 + c o s 2 θ) + s i n 2 θ s i n θ] = \frac{- 1}{2} [c o s θ + c o s 2 θ c o s θ + s i n 2 θ s i n θ] = \frac{- 1}{2} [c o s θ + c o s θ] = - c o s θ ∴ - 1 < - c o s θ < 1 \Rightarrow a \in (- 1, 1)$

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