az^2 + bz + c = 0 problem

[Math790]

If anyone could help me with this problem it would be much apperciated.

Find the equation az^2+ bz + c =0 such that a=1 and the solutions are
3(cos π/3 + i sin π/3) and 2(cos 5π/6 + i sin 5π/6).

 

[Steven G]

If anyone could help me with this problem it would be much apperciated.

Find the equation az^2+ bz + c =0 such that a=1 and the solutions are
3(cos π/3 + i sinπ/3) and 2(cos5π/6 + i sin 5π/6).

If z=r and z=s are roots of an equation then the equation you want is a(z-r)(z-s)=0 (for all values for a). If you are told that a=1, fine, then a=1
You are given two roots. So the equation is....

 

[Math790]

Thank you for the reply. Sadly I'm either getting lost on your explanation or have explained something wrong. I understand r = radius but however don't understand what "s" means.

Also I should clarify the "z" formula.

a +bi = z
r(cos +i sin) =z
r(cos)+r(i sin) = z

When I turn both of the given complex equations from polar to rectangular form neither of them gives me a=1. I think I'm simply going at this problem from the wrong way.

 

[Steven G]

Thank you for the reply. Sadly I'm either getting lost on your explanation or have explained something wrong. I understand r = radius but however don't understand what "s" means.

Also I should clarify the "z" formula.

a +bi = z
r(cos +i sin) =z
r(cos)+r(i sin) = z

When I turn both of the given complex equations from polar to rectangular form neither of them gives me a=1. I think I'm simply going at this problem from the wrong way.

But I defined r and s. They are roots of the equation.

Here is a simpler question. If x=3 and x=4 are the only roots of an equation then what is the equation?

In Pre calculus you should have learned that (linear) factors give roots and roots give factors.

1st answer my question above and we'll go from there.

 

[Math790]

Here is a simpler question. If x=3 and x=4 are the only roots of an equation then what is the equation?

(x-4)(x-3) = 0

going by this logic would it be?

(z- 3cis(π/3)) (z - 2cis(5π/6)) = 0

 

[Steven G]

(x-4)(x-3) = 0

going by this logic would it be?

(z- 3cis(π/3)) (z - 2cis(5π/6)) = 0

This would work. My only concern is that your instructor might want you to go a bit further.

[z-3(cos π/3 + i sin π/3)][z-2(cos 5π/6 + i sin 5π/6) ]=0

z^2 - [3(cis π/3)+2(cis 5π/6]z +6(cis π/3)(cis 5π/6)

Try to use some trig identities especially in the last term after multiply out

 

[Math790]

z^2 - [3(cis π/3)+2(cis 5π/6]z +6(cis π/3)(cis 5π/6)

Try to use some trig identities especially in the last term after multiply out

You are right I could simplify the last term.

z^2 - [3(cis π/3)+2(cis 5π/6]z +6(cis 7π/6)

Now the equation is in the right format

a = 1.
az^2 = (1)z^2
1(z)^2 = z^2

Thank you for all your help.

 

[HallsofIvy]

If a= 0 and \(\displaystyle 3(cos(\pi/3)+ i sin(\pi/3))\) and \(\displaystyle 2(cos(5\pi/3)+ i sin(5\pi/3)\) then the equation must be of the form \(\displaystyle (x- 3(cos(\pi/3)+ i sin(\pi/3))(x- 2(cos(5\pi/3)+ i sin(5\pi/3))\)

Perhaps what is bothering you, is that this has complex coefficients. That has to be true. In order that a polynomial has real coefficients, if a+ bi is a root, then its complex conjugate, a- bi, must also be a root, That is not the case here.