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If `a`, `b`, `c` `p`, `q`, `r` six complex numbers such that `p/a+q/b+r/c=1+i` and `a/p+b/q+c/r=0` then value of `p^2/a^2+q^2/b^2+r^2/c^2=` |
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Answer» `p^2/a^2+q^2/b^2+r^2/c^2 = (p/a+q/b+r/c)^2-2((pq)/(ab)+(qr)/(bc)+(rp)/(ca))` `=(1+i)^2-2((pqc)/(abc)+(qra)/(abc)+(rpb)/(cab))` `=(1+i)^2-2/(abc)*(pqr)(c/r+a/p+b/q)` As, `(a/p+b/q+c/r) = 0`, our equation becomes, `p^2/a^2+q^2/b^2+r^2/c^2 = (1+i)^2` `=1+i^2+2i = 1-1+2i = 2i` `:. p^2/a^2+q^2/b^2+r^2/c^2 = 2i` |
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