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Question

If |z|1,|w|1, then |zw|2(|z||w|)2+(argzargw)2. If this is true enter 1, else enter 0.

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Solution

From the figure
α=(argzargω) ....... (1)
and sin2α/2(α/2)2 ...... (2)
In ΔOAB, from cosine rule
(AB)2=(Oa)2+(OB)22OA.OBcosα
|zω|2=|z|2+|ω|22|z||ω|cosα
|zω|2=(|z||ω|)2+2|z||ω|(1cosα)
|zω|2=(|z||ω|)2+4|z||ω|sin2α/2
|zω|2(|z||ω|)2+|z||ω|α2 [from (2)]
|zω|2(|z||ω|)2+α2 {|z|1,|ω|1}
|zω|2(|z||ω|)2+(argzargω)2 {from (1)}

Ans: 1
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