If z1 and z2 are two non-zero complex numbers such that ∣∣z2z1 ∣∣=2 and arg((z1z2)=23π , then z2z1ˉ is equal

Let z1=a+ib and z2=c+id∣∣z2z1∣∣=2∣z1∣=2∣z2∣arg(z1z2)=23πtan−1ab+tan−1cd=23πtan−1ad−bcbd+ac=23πso,ad=bcab=cdfrom ∣z1∣=2∣z2∣a2+b2=4(c2+d2)ca=2so z2z1=c+ida+ib=(ca)1+idc1+iabca=2so option′A′ is correct.

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If {z}_{1} and {z}_{2} are two non-zero complex numbers such t

Question

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ ∣ \frac{z _{1}}{z _{2}} ∣ ∣ = 2$ and $a r g ((z_{1} z_{2}) = \frac{3 π}{2}$ , then $\frac{z _{1} ˉ}{z _{2}}$ is equal

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Eleanor Discussed

z_{1}

and

z_{2}

are two non-zero complex numbers such that

∣ ∣ \frac{z _{1}}{z _{2}} ∣ ∣ = 2

and

a r g ((z_{1} z_{2}) = \frac{3 π}{2}

, then

\frac{z _{1} ˉ}{z _{2}}

is equal

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Text solutionVerified

Let

z_{1} = a + ib

and

z_{2} = c + i d

∣ ∣ \frac{z _{1}}{z _{2}} ∣ ∣ = 2

∣ z_{1} ∣ = 2 ∣ z_{2} ∣

arg(

z_{1} z_{2}) = \frac{3 π}{2}

tan^{- 1} \frac{b}{a} + tan^{- 1} \frac{d}{c} = \frac{3 π}{2}

tan^{- 1} \frac{b d + a c}{a d - b c} = \frac{3 π}{2}

so,

a d = b c

\frac{b}{a} = \frac{d}{c}

from

∣ z_{1} ∣ = 2 ∣ z_{2} ∣

a^{2} + b^{2} = 4 (c^{2} + d^{2})

\frac{a}{c}

\frac{z _{1}}{z _{2}} = \frac{a + ib}{c + i d} = (\frac{a}{c}) \frac{1 + i \frac{b}{a}}{1 + i \frac{c}{d}}

\frac{a}{c}

so option

^{'} A^{'}

is correct.

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Question Text	If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ ∣ \frac{z _{1}}{z _{2}} ∣ ∣ = 2$ and $a r g ((z_{1} z_{2}) = \frac{3 π}{2}$ , then $\frac{z _{1} ˉ}{z _{2}}$ is equal
Answer Type	Text solution:1
Upvotes	150

If z1​ and z2​ are two non-zero complex numbers such that ∣∣​z2​z1​​ ∣∣​=2 and arg((z1​z2​)=23π ​, then z2​z1​ˉ​ ​ is equal

Text solutionVerified

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If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ ∣ \frac{z _{1}}{z _{2}} ∣ ∣ = 2$ and $a r g ((z_{1} z_{2}) = \frac{3 π}{2}$ , then $\frac{z _{1} ˉ}{z _{2}}$ is equal