An easy function problem.

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Suppose there exists a function such that Q ( x , y ) = x ⋅ Q ( x , 1 ) + y ⋅ Q ( 1 , y ) + x y . Q(x,y) = x\cdot Q(x,1) + y\cdot Q(1,y) + xy. Allow for i = − 1 i=\sqrt{-1} . The value of Q ( 7 i , Q ( 5 , 3 i ) ) Q(7i,Q(5,3i)) can be expressed in the form a + b i a+bi ; determine ∣ a + b ∣ − 13 |a+b|-13 .


The answer is 120.

D S by D S , 31, USA

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1 solution

Tianbo Chen
Jan 13, 2014

First we notice that,

Q ( 1 , 1 ) = Q ( 1 , 1 ) + Q ( 1 , 1 ) + 1 ⇒ Q ( 1 , 1 ) = − 1 Q(1,1) = Q(1,1) + Q(1,1) + 1\Rightarrow Q(1,1) = -1

and

Q ( x , 1 ) = x Q ( x , 1 ) + Q ( 1 , 1 ) + x ⇒ ( Q ( x , 1 ) + 1 ) ( x + 1 ) = 0 Q(x,1) = xQ(x,1) + Q(1,1) +x \Rightarrow (Q(x,1) +1)(x+1) = 0

From the second and first equation we see Q ( x , 1 ) = − 1 Q(x,1) = -1 for all x. Similarly, we can do the same for y and Q ( 1 , y ) = − 1 Q(1,y) = -1 for all y. Therefore Q ( x , y ) = − x − y + x y Q(x,y) = -x -y + xy Plugging in the given values we get − 79 − 54 i ⇒ 79 + 54 − 13 = 120 -79-54i \Rightarrow 79+54-13 = \boxed{120}

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