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Question

If arg (Z1) = α , arg (Z2) = β, the value of |Z1+Z2|2 is


A

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B

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C

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Solution

The correct option is C


We know that |Z|)2 = z¯¯¯z

|Z1+Z2|2 = (|z1+z2|) ¯(|z1+z2|2)

= |Z1+Z2|2 (¯¯¯¯¯z1+¯¯¯¯¯z2) [¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯z1+z2=¯¯¯¯¯z1+¯¯¯¯¯z2]

= z1 ¯¯¯¯¯z1 + z1 ¯¯¯¯¯z2 + z2 ¯¯¯¯¯z1 + z2 ¯¯¯¯¯z2

= |Z1|2 + |Z2|2 + z1 ¯¯¯¯¯z2 + z2 ¯¯¯¯¯z1-------------------(1)

Let arg (Z1) = α, arg(Z2) = β

z1 = |Z1| (cos α + i sinα)

z2 = |Z2| (cos β + i sinβ)

So, z1 ¯¯¯¯¯z2 + z2 ¯¯¯¯¯z1 = |z1|.|z2|.(cos α + i sinα) (cos β - i sinβ) + |z1|.|z2|.(cos α - i sinα) (cos β + i sinβ) [arg(¯¯¯z)=argz]

= |z1|.|z2| [cosα.cosβ + i (-cosα.sinβ + sinα.cosβ) + sinα.sinβ] + [cosα.cosβ + i (cosα.sinβ - sinα.cosβ) + sinα.sinβ]

= 2|z1|.|z2| [cosα.cosβ + sinα.sinβ] = 2|z1|.|z2| = cos(αβ)---------(2)

From (1) and (2),

|Z1+Z2|2 = |z1|2 + |z2|2 + 2|z1|.|z2| cos(αβ)

Where α = arg(z1)

β = arg(z2)


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