106(i know congruencies not der but remainders still can come)..
but still wanna know if (c) will be correct or not
Para:
Let a,b,c are complex numbers and the roots of the equation z3+az2+bz+c=0 are unimodular
Q1. The value of |a|-|b| = (ans got =0)
Q2. The number of distinct real roots of the equation
z3+|a|z2+|b|z+|c|=0 is
(a) 1 (b) 2 (c) 3 (d) NOT
Q3. All roots of z3+|a|z2+|b|z+|c|=0 satisfy
(a) |z| < 1
(b) |z| =1
(c) |z| > 1
(d) NOT
adding a separate question:
MULTI ANSWER
If N=144255+192255, then N is divisible by
(a) 7 (b) 35 (c) 49 (d) 28
i got (a) and (d)
definitely not (b)
just worried abt (c) (as ans given acd)
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10 Answers
106
106sir, for which one?
the remainders one?
if so, then i just needed a Y/N (to confirm that fjee wrong)
341Q 2) |c| = 1 is easy.
So the equation is z^3+|a|z^2 + |a| z + 1 = 0.
If all roots are real then we must have |a|^2 \ge 3|a| \Rightarrow |a| \ge 3
(This because of the well known inequality (\sum p)^2 \ge 3 \sum pq
But using sum of roots relation
|a| = |\alpha + \beta + \gamma| \le |\alpha| + |\beta| + |\gamma| \le 3
If |a| =3, z=1 is the only repeated root.
If |a|<3, then we have z=-1 as the only real root
341ya, that remainder wala.
Just for completeness, 144^{255} + 196{255} \equiv -(3^{255} + 4^{255}) \pmod {49}
(shud be 196255)
We then use Euler Totient Theorem that if (a,49) = 1 then
a^{42} \equiv 1 \pmod {49} \Rightarrow a^{255} \equiv a^3 \pmod{49}
From here its easy to finish
341For 3, now that we know that z=-1 is a root
So for the other two non-real roots (thats when |a|<3), we have |\beta \gamma| = 1
But they are conjugates so |\beta| = |\gamma| =1
Option B
edited: a tubelight came on!