# Complex numbers with a polenom!

• Mar 19th 2011, 01:56 AM
Complex numbers with a polenom!
Hello,

So here is this combined polynomial & complex numbers; I couldn't solve it!

F(z)= z^7 + a(6)*z^6 + a(5)*z^5 + a(4)*z^4 + a(3)*z^3 + a(2)*z^2 + a(1)*z^1+a(0)

It is given that z=-4-2i is a root of F(z),F'(z),F''(z) (I guess it's correct to say that z=-4+2i is also root for the three functions!)

a(0)=-6

a(0),,,,,,a(6) are all real numbers

What is the value of a(6) ?

**a(Number) indicates for index!

Any thoughts?
• Mar 19th 2011, 03:04 AM
Ackbeet
Here's one thought: if the coefficients of the polynomial are all real, then any complex roots come in complex-conjugate pairs. Hence, -4+2i is also a root of F, F', and F''. That means you technically have 7 equations and 7 unknowns (the unknowns being the coefficients). I would just start plugging in numbers into the functions you have, and start cranking away. It's a fair bit of algebra, looks like. What do you get?
• Mar 19th 2011, 03:27 AM
Quote:

Originally Posted by Ackbeet
Here's one thought: if the coefficients of the polynomial are all real, then any complex roots come in complex-conjugate pairs. Hence, -4+2i is also a root of F, F', and F''. That means you technically have 7 equations and 7 unknowns (the unknowns being the coefficients). I would just start plugging in numbers into the functions you have, and start cranking away. It's a fair bit of algebra, looks like. What do you get?

Are you sure this is a good was to solve this it mayet take me hours
• Mar 19th 2011, 06:33 AM
Ackbeet
Are you allowed to use a Computer Algebra System like Mathematica?
• Mar 19th 2011, 07:01 AM
awkward
Since -4-2i is a zero of F, F', and F'', (z + 4 + 2i)^3 must be a factor of F. The same is true of -4+2i, so (z + 4 -2i)^3 is also a factor. Let's say the remaining zero of F is a. Then

F(z) = (z-a) (z + 4 + 2i)^3 (z + 4 -2i)^3
= (z-a) [(z + 4 + 2i) (z + 4 -2i)]^3
= (z-a) (z^2 + 8z + 20)^3

You can use F(0) = -6 to determine a. So...
• Mar 19th 2011, 07:20 AM
Ackbeet
Quote:

Originally Posted by awkward
Since -4-2i is a zero of F, F', and F'', (z + 4 + 2i)^3 must be a factor of F. The same is true of -4+2i, so (z + 4 -2i)^3 is also a factor. Let's say the remaining zero of F is a. Then

F(z) = (z-a) (z + 4 + 2i)^3 (z + 4 -2i)^3
= (z-a) [(z + 4 + 2i) (z + 4 -2i)]^3
= (z-a) (z^2 + 8z + 20)^3

You can use F(0) = -6 to determine a. So...

• Mar 19th 2011, 08:22 AM
Hello,first thank u all,

second off "Computer Algebra System" please if u got a link to such a calculator then post it,

second the method with

F(z) = (z-a) (z + 4 + 2i)^3 (z + 4 -2i)^3
= (z-a) [(z + 4 + 2i) (z + 4 -2i)]^3
= (z-a) (z^2 + 8z + 20)^3

You can use F(0) = -6 to determine a. So...

I put z=0 then I get (-a)[(4+2i)(4-2i)]^3 = (-a)(20)^3

this equals to -6 => a*20^3=6 => a=6/(20^3)

then what?, I gotta find a(6) to open operators is bit hard!
• Mar 19th 2011, 10:23 AM