Question

Q.If a,b,c,q,r are six complex numbers,such that p/a+q/b+r/c=1+i and a/p+b/q+c/r=0,where i=root(-1),then the value of p^2/a^2+q^2/b^2+r^2/c^2?


anyone please answer this question if you know ☺don't unnecessary answer please.​

Answer 1

Answer:2

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Step-by-step explanation:

p2/a2+q2/b2+r2/c2=(pa+qb+rc)2−2(pqab+qrbc+rpca)

=(1+i)2−2(pqcabc+qraabc+rpbcab)

=(1+i)2−2abc⋅(pqr)(cr+ap+bq)

As, (ap+bq+cr)=0, our equation becomes\\

p2\a2+q2/b2+r2/c2=(1+i)2

=1+i2+2i=1−1+2i=2i

∴p2a2+q2/b2+r2/c2=2

Answer 2

Question:-

If $\displaystyle\sf {a,\ b,\ c,\ p,\ q,\ r}$ are six complex numbers such that,

$\displaystyle\sf {\dfrac {p}{a}+\dfrac {q}{b}+\dfrac {r}{c}=1+i}$

and,

$\displaystyle\sf {\dfrac {a}{p}+\dfrac {b}{q}+\dfrac {c}{r}=0}$

where $\displaystyle\sf {i=\sqrt{-1}}$, then find the value of,

$\displaystyle\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}.}$

$\quad$

Answer:-

$\quad$

$\displaystyle\large\boxed {\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=2i}}$

$\quad$

Solution:-

$\quad$

Well,

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=\left (\dfrac {p}{a}+\dfrac {q}{b}+\dfrac {r}{c}\right)^2-2\left (\dfrac {pq}{ab}+\dfrac {qr}{bc}+\dfrac {pr}{ac}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=(1+i)^2-2\left (\dfrac {pqc}{abc}+\dfrac {aqr}{abc}+\dfrac {pbr}{abc}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=(1+i)^2-\dfrac {2}{abc}\left (\dfrac {pqrc}{r}+\dfrac {apqr}{p}+\dfrac {pqbr}{q}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=(1+i)^2-\dfrac {2pqr}{abc}\left (\dfrac {a}{p}+\dfrac {b}{q}+\dfrac {c}{r}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=(1+i)^2-\dfrac {2pqr}{abc}\left (0\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=(1+i)^2}$

$\quad$

$\displaystyle\longrightarrow\sf {\underline {\underline {\dfrac {p^2}{a^2}+\dfrac {q^2}{b^2}+\dfrac {r^2}{c^2}=2i}}}$