SOLUTION: z1 and z2 are two complex numbers. if |z1+z2|=|z1|+|z2| then show that arg(z1)=arg(z2)

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Question 16100: z1 and z2 are two complex numbers. if |z1+z2|=|z1|+|z2| then show that arg(z1)=arg(z2)
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
|z1+z2|=|z1|+|z2|
(|z1+z2|) ^2 = (|z1|+|z2| )^2 = |z1|^2+|z2|^2+2|z1||z2|
let z1=x1+iy1
z2=x2+iy2
hence
( |x1+iy1+x2+iy2| )^2 = ( |x1+iy1| )^2+( |x2+iy2| )^2 +2( |x1+iy1| )( |x2+iy2| )
( |(x1+x2)+i(y1+y2)| ) ^2 = (x1^2+y1^2) +(x2^2+y2^2)+2 sqrt%28%28x1%5E2%2By1%5E2%29%28x2%5E2%2By2%5E2%29%29
(x1+x2)^2+(y1+y2)^2 = (x1^2+y1^2) +(x2^2+y2^2)+2 sqrt%28%28x1%5E2%2By1%5E2%29%28x2%5E2%2By2%5E2%29%29
x1^2+x2^2+2x1*x2+y1^2+y2^2+2y1*y2 = (x1^2+y1^2) +(x2^2+y2^2)+2 sqrt%28%28x1%5E2%2By1%5E2%29%28x2%5E2%2By2%5E2%29%29
2(x1*x2+y1*y2) = 2 sqrt%28%28x1%5E2%2By1%5E2%29%28x2%5E2%2By2%5E2%29%29
cancelling 2 on either side and squaring both sides
(x1x2+y1y2)^2 = (x1^2+y1^2)(x2^2+y2^2)
(x1^2)(x2^2)+(y1^2)(y2^2) +2x1*x2*y1*y2 = (x1^2)(x2^2)+(y1^2)(y2^2)+(x1^2)(y2^2)+(x2^2)(y1^2)
(x1^2)(y2^2)+(x2^2)(y1^2) - 2x1*x2*y1*y2 = 0
(x1y2-x2y1)^2 = 0
x1y2 - x2y1 = 0
x1y2 = x2y1
y1/x1 = y2/x2
Tan (y1/x1) = Tan (y2/x2)
Hence argument of z1 = argument of z2