1forget it .. it wont cum in exam
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(i know congruencies not der but remainders still can come)..
but still wanna know if (c) will be correct or not
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!