Question

A polynomial 2z^5+az^4+bz^3+cz^2+dz+e with real coefficients a,b,d,e has 2i, -i and 1 as roots. Find c.

Answer 1

\[2z^5+az^4+bz^3+cz^2+dz+e = 2(z-1)(z+2i)(z-2i)(z+i)(z-i)\]

Answer 2

compare z^2 coefficient both sides

Answer 3

maybe simplify the right hand side a bit first using : (a+bi)(a-bi) = a^2 + b^2

Answer 4

not sure if there is any other easy way..

Answer 5

Lol, I hope to God there is xD

Answer 6

hmm there could be, but i feel its not worth trying for other ways as they're not so obvious lol..

Answer 7

i would just expand and compare coefficients both sides :
\[2z^5+az^4+bz^3+cz^2+dz+e = 2(z-1)(z+2i)(z-2i)(z+i)(z-i) \]
\[2z^5+az^4+bz^3+cz^2+dz+e = 2(z-1)(z^2+4)(z^2+1)\]

Answer 8

Ooooooh, I see. So just open it up and see what has z^2.

Answer 9

That's not too bad.

Answer 10

exactly ! its trivial to solve, but agree its a dumb method lol

Answer 11

Thanks x)

Answer 12

One question. Why z+i and z-i?

Answer 13

thats a very good question :)
cuz of a theorem : if a is `k` root of `P(x)`, then `(x-k)` is a factor of `P(x)`

Answer 14

and one more theorem : complex roots always come in conjugate pairs (if `a + bi` is a root, then `a - bi` will also be a root)

Answer 15

But the roots are, 2i, -i, 1. Wouldn't it be: (z-2i) ( z+i) (z-1)?

Answer 16

Oh, I see.

Answer 17

:)

Answer 18

That helps a lot. Gonna be a problem on my final tomorrow. Have a good weekend :)

Answer 19

if you're feeling lazy, ur z^2 coefficient is here :
http://www.wolframalpha.com/input/?i=simplify+2%28z-1%29%28z%2B2i%29%28z-2i%29%28z%2Bi%29%28z-i%29

Answer 20

Hahaha, no wolfram on exams x)

Answer 21

wish u all the best wid the exam :)

Answer 22

thanks :)