Complex Numbers (1 Viewer)

[cookiez69]

Hey guys,

Given that a and b are real, and that (2+i) is a root of the equation
z^2 + (a+9i)z + b(2+11i) = 0

Determine the values of a and b.

Thanks

 

[Carrotsticks]

Hey guys,

Given that a and b are real, and that (2+i) is a root of the equation
z^2 + (a+9i)z + b(2+11i) = 0

Determine the values of a and b.

Thanks

Since 2+i is a root, P(2+i) must be equal to zero.

So sub it in and let it equal zero, then separate imaginary and real components, and you should get (big real component with a's and b's) + i(big imaginary component with a's and b's) = 0

But 0 = 0+0i, so equate real and imaginary components to form a set of simultaneous equations, and then solve for a and b.

======================

Another way is to let the other root be A, so using sum and product of roots you have A(2+i) = b(2+11i), so A=b(2+11i)/(2+i). Form a similar expression using sum of roots and make A the subject, then substitute into our first equation.

Then you should get this big equation in terms of a and b. Separate real/imaginary components and chuck it all on one side, then proceed similarly to the above method where you equate real and imaginary components with 0=0+0i etc...

 

[HeroicPandas]

In ur 2nd method Carrot, A can be (2-i) because of the fantastic complex conjugate root theorem!

Since (2+i) is a complex root, then its complex conjugate (2-i) is also a root

 

[RealiseNothing]

In ur 2nd method Carrot, A can be (2-i) because of the fantastic complex conjugate root theorem!

Since (2+i) is a complex root, then its complex conjugate (2-i) is also a root

The co-efficients of the quadratic are not real.

 

[HeroicPandas]

The co-efficients of the quadratic are not real.

damnn woops lol

 

[asianese]

It's a pretty fantastic theorem, though