Let z 1 be a complex number and let - - x - iy 0 - If - - - z1-z is purely real- then -z- is equal to -

The correct option is D 1 Given, ω = x+iy and ω ωz1 z = R (Real) z = R ωR ω z = R x iyR x+iy = (R x)^2 y^2 + 2(R x)y i(R x)^2+y^2 | ...

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

Byju's Answer

Standard XII

Mathematics

Complex Numbers

Let z 1 be ...

Question

Let $z \neq 1$ be a complex number and let $ω = x + i y \neq 0 .$ If $\frac{ω - ω z}{1 - z}$ is purely real, then $| z |$ is equal to :

| ω |

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

| ω |^{2}

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

\frac{1}{| ω |^{2}}

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

\frac{1}{| ω |}

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

1

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

Open in App

Solution

The correct option is D $1$
Given, $ω = x + i y$ and $\frac{ω - ¯ ¯ ¯ ω z}{1 - z} = R (R e a l) ⟹ z = \frac{R - ω}{R - ¯ ¯ ¯ ω}$

$⟹ z = \frac{R - x - i y}{R - x + i y} = \frac{(R - x)^{2} - y^{2} + 2 (R - x) y i}{(R - x)^{2} + y^{2}}$

$⟹ | z | = \frac{1}{(R - x)^{2} + y^{2}} \sqrt{{(R - x)^{2} - y^{2}}^{2} + {2 (R - x)}^{2}} = \frac{(R - x)^{2} + y^{2}}{(R - x)^{2} + y^{2}} = 1$

Suggest Corrections

Similar questions

Q. Let

z

and

ω

two complex number such that

| z | \leq 1, | ω | \leq 1

and

| z - i ω | = | z - i ¯ ¯ ¯ ω | = 2

, then

z

equals to

Q. Let

z

and

ω

be two complex numbers such that

| z | \leq 1, | ω | \leq 1

and

| z + i ω | = | z - i ¯ ω | = 2

, then

z

equals

Q. Let

Z = x + i y

and

ω = \frac{1 - i Z}{Z - i}

. If

| ω | = 1

, show that

Z

is purely real.

Q. Let z and

ω

be two complex numbers. If Re(z) =

| z - 2 |, R e (ω) = | ω - 2 |

and arg (z

-

ω

) =

\frac{π}{3}

, then Im (z +

ω

Q. Let

z

ω

be two complex numbers such that

| z | = 1, | ω | = 1

and

| z + i ω | = | z - i ¯ ¯ ¯ ω | = 2

, then

z

equals

Join BYJU'S Learning Program

Let z≠1 be a complex number and let ω=x+iy≠0. If ω−ωz1−z is purely real, then |z| is equal to :

The correct option is D 1Given, ω=x+iy and ω−¯¯¯ωz1−z=R(Real)⟹z=R−ωR−¯¯¯ω⟹z=R−x−iyR−x+iy=(R−x)2−y2+2(R−x)yi(R−x)2+y2⟹|z|=1(R−x)2+y2√{(R−x)2−y2}2+{2(R−x)}2=(R−x)2+y2(R−x)2+y2=1

Let $z \neq 1$ be a complex number and let $ω = x + i y \neq 0 .$ If $\frac{ω - ω z}{1 - z}$ is purely real, then $| z |$ is equal to :