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1 Answer

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by (35 points)

Given,

let lz2/mz1 = iy

z2/z1 = iym/l

|lambda z1+µz2/lamda z1-µz2|

   dividing numerator and denominator by z1 we get

|lambda(1)+µ(z2/z1) / lambda(1)-µ(z2/z1)|

    we know that z2/z1 = imy/l 

|lambda+µ(imy/l) / lambda-µ(imy/l)|

    we know that |x+iy| = root of (x2+y2)

⇒square root(lambda² + µ²y²/l² )/square root( lambda² + µ²y²/l²)

1

hope this helps :)

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