If - z -1 and z - 1- then all the values of z -1- z 2 lie onA- The X - axisB- The Y - axisC- a line not passing through the originD- - z -2

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(cosθ+i sinθ) =i/2sinθ Hence, z/1 z^2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

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 $\frac{z}{1 - z^{2}}$ lie on

a

line not passing through the origin

No worries! We‘ve got your back. Try BYJU‘S free classes today!

| z | = \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

The X - axis

No worries! We‘ve got your back. Try BYJU‘S free classes today!

The Y - axis

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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.

Suggest Corrections

Similar questions

Q. If

| z | = 1

and

z \neq \pm 1

, then all the values of

\frac{z}{1 - z^{2}}

lie on

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

Q. The complex numbers

z = x + i y

which satisfy the equation

∣ \frac{z - 5 i}{z + 5 i} ∣= 1

lie on

Q. If

| z^{2} - 1 | = | z |^{2} + 1

, then

z

lies on

Q. If

z = x + i y

is a complex number which satisfy

\frac{| z - 5 i |}{| z + 5 i |} = 1

, then

z

lies on

Join BYJU'S Learning Program

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

The correct option is D The Y - axisLet 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 $\frac{z}{1 - z^{2}}$ lie on