If - z -1 and Z - 1- then all the values of z -1- z 2 lies onA- The Y - axisB- a line not passing through the originC- - z -2D- The X - axis

The correct option is A The Y axis Let z=cosθ + i sinθ ⇒z/1 z^2=cosθ +sinθ/1 (cos2θ +i sin2θ) =cosθ +i sinθ/2sin^2 θ 2i sinθ cosθ =cosθ +i sinθ/ 2i sinθ( ...

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Byju's Answer

Standard XI

Mathematics

Polar Representation of a Complex Number

If | z |=1 an...

Question

If $| z | = 1$ and $z \neq \pm 1$ , then all the values of $\frac{z}{1 - z^{2}}$ lies on

$a$ line not passing through the origin

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$| z | = \sqrt{2}$

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The X - axis

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The Y - axis

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Solution

The correct option is D
The Y - axis

Let $z = c o s θ + i s i n θ \Rightarrow \frac{z}{1 - z^{2}} = \frac{c o s θ + s i n θ}{1 - (c o s 2 θ + i s i n 2 θ)} = \frac{c o s θ + i s i n θ}{2 s i n^{2} θ - 2 i s i n θ c o s θ} = \frac{c o s θ + i s i n θ}{- 2 i s i n θ (c o s θ + i s i n θ)} = \frac{i}{2 s i n θ}$
Hence, $\frac{z}{1 - z^{2}}$ lies on the imaginary axis i.e. Y-axis.

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If |z|=1 and z≠±1, then all the values of z1−z2 lies on

The correct option is D The Y - axis Let z=cosθ+i sinθ⇒z1−z2=cosθ+sinθ1−(cos2θ+i sin2θ)=cosθ+i sinθ2sin2θ−2i sinθcosθ=cosθ+i sinθ−2i sinθ(cosθ+i sinθ)=i2sinθ Hence, z1−z2 lies on the imaginary axis i.e. Y-axis.

If $| z | = 1$ and $z \neq \pm 1$ , then all the values of $\frac{z}{1 - z^{2}}$ lies on