Z1- Z2 are two points in an argand plane- If a-Z1- - b-Z1-Prove that aZ1 - bZ2aZ1 - bZ2 is purely imaginary-

Let Z_1 = r_1 e^iθ, #160;Z_2 = r_2 e^i(θ+α) .Given that ar_1 = br_2 Z = aZ_1 bZ_2aZ_1+bZ_2 = #160;e^iθ e^i(θ+α)e^iθ + e^i(θ + α) = 1 ...

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

Geometrical Representation of a Complex Number

Z1≠ Z2 are tw...

Question

$Z_{1} \neq Z_{2}$ 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.

Open in App

Solution

Suggest Corrections

Similar questions

| z_{1} |

| z_{2} |

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

a, b \in R

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

z_{1}

z_{2}

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

ϵ

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

z_{1}

z_{2}

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

a, b ϵ R

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

(z_{1}) = θ

(z_{2}) = θ + α

\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}

z_{2}

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

a, b \in R)

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

z_{1}

z_{2}

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

a, b \in R),

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

Join BYJU'S Learning Program