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If `((a+i)^2)/((2a-i))=p+i q ,`show that: `p^2+q^2=((a^2+1)^2)/((4a^2+1))`. |
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Answer» `(a+i)^2/(2a-i) = p+iq->(1)` `:. bar((a+i)^2/bar(2a-i)) = p-iq` `=>(a-i)^2/(2a+i) = p-iq->(2)` Multiplying (1) and (2), `((a+i)^2(a-i)^2)/((2a-i)(2a+i)) = (p+iq)(p-iq)` `=>((a+i)(a-i))^2/(4a^2 - i^2) = p^2-i^2q^2` `=>(a^2-i^2)^2/(4a^2+1) = p^2+q^2 ...[As i^2 = -1]` `=>(a^2+1)^2/(4a^2+1) = p^2+q^2` |
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