If z 1 and z 2 are two distinct points in an argand plane such that a - z 1- b - z 2- where a - b - R - then the point az 1- bz 2- bz 2- az 1 will always lie on theA- line segment -2-2- of the real axisB- line segment -2-2- of the imaginary axisC- unit circle -z-1D- the line whose z-tan -1 2

Assuming (z_1) = θ and (z_2) = θ + α, az_1bz_2 + bz_2az_1 = a|z_1|e^i θb|z_2|e^i(θ + α) + b|z_2|e^i(θ + α)a |z_1|e^i θ = e^ i α + e^i α = 2 cosα Hence, the poin ...

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Byju's Answer

Standard XII

Mathematics

Rotation Concept

If z 1 and z ...

Question

If $z_{1}$ and $z_{2}$ are two distinct points in an argand plane such that $a | z_{1} | = b | z_{2} |$ (where $a, b \in R)$ , then the point $(\frac{a z_{1}}{b z_{2}} + \frac{b z_{2}}{a z_{1}})$ will always lie on the

line segment

[- 2, 2]

of the real axis

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line segment

[- 2, 2]

of the imaginary axis

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unit circle

| z | = 1

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the line whose

arg (z) = {tan}^{- 1} 2

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Solution

The correct option is A line segment $[- 2, 2]$ of the real axis
Assuming $arg (z_{1}) = θ$ and $arg (z_{2}) = θ + α,$
$\frac{a z_{1}}{b z_{2}} + \frac{b z_{2}}{a z_{1}} = \frac{a | z_{1} | e^{i θ}}{b | z_{2} | e^{i (θ + α)}} + \frac{b | z_{2} | e^{i (θ + α)}}{a | z_{1} | e^{i θ}}$
$= e^{- i α} + e^{i α} = 2 cos α$
Hence, the point will always lie on the line segment $[- 2, 2]$ of the real axis.

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Similar questions

Q. Assertion :Let

z_{1}

z_{2}

are two distinct points in the argand plane such that

a | z_{1} | = b | z_{2} |

where

a, b ϵ R

.
The expression

\frac{a z_{1}}{b z_{2}} + \frac{b z_{2}}{a z_{1}}

is a point on the line segment [-2, 2] of the real axis. Reason: When arg

(z_{1}) = θ

& arg

(z_{2}) = θ + α

, then

\frac{a z_{1}}{b z_{2}} + \frac{b z_{2}}{a z_{1}} = e^{i α} + e^{- i α} = 2 c o s α

| z_{1} |

and

| z_{2} |

are two distinct points in an Argand plane. If

a | z_{1} | = b | z_{2} |

(where

a, b \in R

), then the point

(a z_{1} / b z_{2}) + (b z_{2} / a z_{1})

is a point on the

Z_{1} \neq Z_{2}

are two points in an argand plane. If

a | Z_{1} | = b | Z_{1} |

Prove that

\frac{a Z_{1} - b Z_{2}}{a Z_{1} + b Z_{2}}

is purely imaginary.

z_{1}

and

z_{2}

are two distinct points in an Argand plane. If

a | z_{1} | = b | z_{2} |

(where a, b

ϵ

R), then the point

(a z_{1} / b z_{2}) + (b z_{2} / a z_{1})

is a point on the

z_{1}

and

z_{2}

are two distinct points in an Argand plane. If

a | z_{1} | = b | z_{2} |

(where

a, b \in R),

then the point

(a z_{1} / b z_{2}) + (b z_{2} / a z_{1})

is a point on the

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If z1 and z2 are two distinct points in an argand plane such that a|z1|=b|z2| (where a,b∈R), then the point (az1bz2+bz2az1) will always lie on the

The correct option is A line segment [−2,2] of the real axisAssuming arg(z1)=θ and arg(z2)=θ+α, az1bz2+bz2az1=a|z1|eiθb|z2|ei(θ+α)+b|z2|ei(θ+α)a|z1|eiθ =e−iα+eiα=2cosα Hence, the point will always lie on the line segment [−2,2] of the real axis.

If $z_{1}$ and $z_{2}$ are two distinct points in an argand plane such that $a | z_{1} | = b | z_{2} |$ (where $a, b \in R)$ , then the point $(\frac{a z_{1}}{b z_{2}} + \frac{b z_{2}}{a z_{1}})$ will always lie on the