Consider a network Q consisting of 2015 distinct parallel impedances (in ohms)
, such that each of these individual values satisfy the equation
.
Determine the total impedance (in ohms) of an impedance network consisting of 2015 Q networks in series.
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It is obvious that Z 1 , Z 2 , . . . Z 2 0 1 5 are zeroes of the polynomial y = ( x − 2 ) 2 0 1 5 − 2 2 0 1 7 . Thus, by Vieta's formulas, we can get the net impedance of network Q, which is the quotient of the constant over the coefficient in x (Do you see why?). That is, 2 0 1 5 ⋅ 2 2 0 1 4 2 2 0 1 5 + 2 2 0 1 7 , which simplifies to 4 0 3 2 ohms. Since there are 2 0 1 5 of these in series, we finally get the answer 2 0 1 5 ⋅ 4 0 3 2 = 1 0 .