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Question : 8 of 160
Marks:
+1,
-0
Solution:
z1=eia,z2=eip,z3=eix ‌=e−iα,‌=e−iβ,‌=e−iγAccording to the questions,
z1+z2+z3=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)Now,
‌+‌+‌ =(cos‌α−isin‌α)+(cos‌β−isin‌β)+(cos‌γ−isin‌γ) =Σ‌cos‌α−iΣsin‌α=0−i⋅0=0 ⇒z1z2+z2z3+z3z1=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) z12+z22+z32=(z1+z2+z3)2−2(z1z2+z2z3+z3z1) ⇒z12+z22+z32=0−0 . [From Eqs. (i) and (ii)]
⇒ei2α+ei2β+ei2γ=0 ⇒cos‌2‌α+cos‌2‌β+cos‌2‌γ=0and
sin‌2α+sin‌2β+sin‌2γ=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)Now,
z1z2+z2z3+z3z1=0 ⇒ei(α+β)+ei(β+γ)+ei(γ+α)=0 ⇒‌‌Σ‌cos(α+β)=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv)and
Σsin‌(α+β)=0From Eqs. (iii) and (iv),
cos‌2‌α+⋅cos‌2‌β+cos‌2‌γ =cos(α+β)+cos(β+γ)+cos(γ+α)
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