Let Z1 and Z2 are two non zero complex numbers such that |Z1+Z2|=|Z1|=|Z2| then (Z1/Z2) can be equal to where ω is the non real cubs root of unity.

Question

Let Z1 and Z2 are two non zero complex numbers such that |Z1+Z2|=|Z1|=|Z2| then (Z1/Z2) can be equal to where ω is the non real cubs root of unity. (A) 1+ω (B) 1+ω2 (C) ω (D) ω2

Tardigrade Admin · Accepted Answer

text Let ( Z 2/ Z 1)= t ⇒ Z 2= Z 1 t , t ≠ 0 ∴ | Z 1(1+ t )|=| Z 1|=| Z 1 t | |1+ t |=1=| t | ⇒ t ⋅ overline t =1 text now (1+ t )(1+ overline t )= t overline t 1+ overline t + t + t overline t = t overline t ∴ 1+ t + overline t =0 1+ t +(1/ t )=0 ⇒ t 2+ t +1=0 t =ω text or ω2 ⇒ C , D

Q. Let $Z_{1}$ and $Z_{2}$ are two non zero complex numbers such that $∣ Z_{1} + Z_{2} ∣ = ∣ Z_{1} ∣ = ∣ Z_{2} ∣$ then $\frac{Z _{1}}{Z _{2}}$ can be equal to
where $ω$ is the non real cubs root of unity.

$1 + ω$

$1 + ω^{2}$

$ω$

$ω^{2}$

Q. Let Z1​ and Z2​ are two non zero complex numbers such that ∣Z1​+Z2​∣=∣Z1​∣=∣Z2​∣ then Z2​Z1​​ can be equal to where ω is the non real cubs root of unity.

1+ω

1+ω2

ω

ω2

Let Z1​Z2​​=t⇒Z2​=Z1​t,t=0 ∴∣Z1​(1+t)∣=∣Z1​∣=∣Z1​t∣ ∣1+t∣=1=∣t∣⇒t⋅t=1 now (1+t)(1+t)=tt 1+t+t+tt=tt ∴1+t+t=0 1+t+t1​=0⇒t2+t+1=0 t=ω or ω2⇒C,D

Q. Let $Z_{1}$ and $Z_{2}$ are two non zero complex numbers such that $∣ Z_{1} + Z_{2} ∣ = ∣ Z_{1} ∣ = ∣ Z_{2} ∣$ then $\frac{Z _{1}}{Z _{2}}$ can be equal to
where $ω$ is the non real cubs root of unity.

$1 + ω$

$1 + ω^{2}$

$ω$

$ω^{2}$