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Question

# If |z1+z2|=|z1−z2| then argz1−argz2=

A
(2n+1)π2,nZ
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B
nπ+π4,nZ
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C
2nπ,nZ
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D
(2n+1)π,nZ
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Solution

## The correct option is A (2n+1)π2,n∈ZGiven that |z1+z2|=|z1−z2| ⇒|z1+z2|2=|z1−z2|2 [∵|z1−z2|2=|z1|2+|z2|2−2Re(z1¯¯¯¯¯z2 ),|z1+z2|2=|z1|2+|z2|2+2Re(z1¯¯¯¯¯z2 )] ⇒|z1|2+|z2|2+2Re(z1¯¯¯¯¯z2 )=|z1|2+|z2|2−2Re(z1¯¯¯¯¯z2 ) ⇒4Re(z1¯¯¯¯¯z2 )=0 ⇒Re(z1¯¯¯¯¯z2 )=0 Let z1=r1eiθ1 and z2=r2eiθ2 Then z1¯¯¯¯¯z2=r1r2eiθ1e−iθ2=r1r2ei(θ1−θ2) ⇒z1¯¯¯¯¯z2=r1r2(cos(θ1−θ2)+isin(θ1−θ2)) ∵Re(z1¯¯¯¯¯z2 )=0 ∴cos(θ1−θ2)=0 ⇒θ1−θ2=(2n+1)π2,n∈Z

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