JEE Advanced 2016 (6)
Say a real number is repetitive if there exist two distinct complex numbers with and is not equal to such that There exist real numbers such that a real number is repetitive if and only if . If the value of can be expressed in the form for relatively prime positive integers . Find .
Try my set JEE ADVANCED 2016
Source: OMO Spring 2016
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1 solution
n. Let the argument of z, with |z| = 1, be θ. Note that f(z) = z
4 + z
3 + rz2 + z + 1 is in the
direction of argument 2θ with a signed magnitude of r + 2 cos θ + 2 cos(2θ) = 4 cos2
θ + 2 cos θ −2 +r =
4(cos θ + 1 4 ) 2 − 9 4 + r. For r > 9 4
, then note that this signed magnitude is always positive. Then, f(z1) = f(z2) can only
happen for z1 6= z2 if their arguments θ1, θ2 satisfy 2θ1 = 2θ2 =⇒ θ1 = θ2 + π. But then their signed
magnitudes can only be equal if cos θ1 = cos θ2, but that implies {z1, z2} = {i, −i}. Hence f(z1) 6= f(z2)
for all distinct z1, z2. For r < −4, the signed magnitude is always negative since | cos θ + 1 4 | ≤ 5 4 over all θ.
Once again, this implies that f(z1) 6= f(z2). However, if −4 ≤ r ≤ 9 4 ,
then there exists a unique value for cos θ with − 1 4 ≤ cos θ ≤ 1 such that the signed magnitude is equal to 0
. For all cos θ 6= 1, there are two values of θ giving the desired value
of cos θ, showing there exist distinct z1, z2 giving f(z1) = f(z2) = 0.
However, cos θ = 1 only when θ = 0, meaning for r = −4 we see that the signed magnitude is always negative except for θ = 0, when the signed magnitude is exactly 0. Hence r = −4 is also repetitive.
Hence a = −4 and b = 9 4 so 100p + q = 2504.