If α,β,γ and a,b,c are complex numbers such that αa+βb+γc=1+i and aα+bβ+cγ=0, then the value of α2a2+β2b2+γ2c2 is equal to _______

αa+βb+γc=1+i⇒α2a2+β2b2+γ2c2+2αβab+βγbc+γαca=2i⇒α2a2+β2b2+γ2c2+2αβγabccγ+aα+bβ=2iα2a2+β2b2+γ2c2+20=2iα2a2+β2b2+γ2c2=2i=2

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If $α, β, γ$ and $a, b, c$ are complex numbers such that $\frac{α}{a} + \frac{β}{b} + \frac{γ}{c} = 1 + i$ and $\frac{a}{α} + \frac{b}{β} + \frac{c}{γ} = 0$ , then the value of $|\frac{α^{2}}{a^{2}} + \frac{β^{2}}{b^{2}} + \frac{γ^{2}}{c^{2}}|$ is equal to _______

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answer is 2.

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Detailed Solution

$\frac{α}{a} + \frac{β}{b} + \frac{γ}{c} = 1 + i \Rightarrow \frac{α^{2}}{a^{2}} + \frac{β^{2}}{b^{2}} + \frac{γ^{2}}{c^{2}} + 2 (\frac{α β}{a b} + \frac{β γ}{b c} + \frac{γ α}{c a}) = 2 i$

$\Rightarrow \frac{α^{2}}{a^{2}} + \frac{β^{2}}{b^{2}} + \frac{γ^{2}}{c^{2}} + 2 \frac{α β γ}{a b c} (\frac{c}{γ} + \frac{a}{α} + \frac{b}{β}) = 2 i$

$\frac{α^{2}}{a^{2}} + \frac{β^{2}}{b^{2}} + \frac{γ^{2}}{c^{2}} + 2 (0) = 2 i$

$|\frac{α^{2}}{a^{2}} + \frac{β^{2}}{b^{2}} + \frac{γ^{2}}{c^{2}}| = |2 i| = 2$

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