Question

b) Prove that ∗ + C S/ , where ~R S z{ |z| }1
*
= ∈C = , = x{ ∈ x| > }0
+ R R and
}0{\
* C = C .b) Prove that ∗ + C S/ , where ~R S z{ |z| }1
*
= ∈C = , = x{ ∈ x| > }0
+ R R and
}0{\
* C = C .

Answer 1

Answer:

Step-by-step explanation:

∴ a + b + c = 0

∴ a = -(b + c), b = -(c + a) , c = -(a + b)

Now, (2a² + bc ) = (a² + a² + bc)

= {a² + a(-b - c) + bc}

= a² - ab - ac + bc = a(a - b) -c(a - c) = (a - c)(a - b) ------------(1)

similarly, (2b² + ca) = (b - c)(b - a) -------------(2)

(2c² + ab) = (c - a)(c - b) --------------(3)

now, LHS = 1/(2a² + bc ) + 1/(2b² + ca) + 1/(2c² + ab)

= 1/(a - b)(a - c) + 1/(b - c)(b - a) + 1/(c - a)(c - b)

= - 1/(a- b)(c - a) -1/(b - c)(a - b) - 1/(c - a)(b - c)

= -[(b -c ) + (c - a) + (a - b)]/(a - b)(b - c)(c - a)]

= 0 = RHS

Hence, proved

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