Any easier/quicker/more correct solutions to this problem (1 Viewer)

[calamebe]

I have a question that I feel i worked out rather weirdly, so I was hoping there was an easier way to do these types of questions. I will post the question and my solution.

Question:



My solution:

http://m.imgur.com/Ej5ZhAt

My solution is correct by the way, just looking for other methods.

 

[dan964]

there is a simpler method
1. This polynomial is of degree 2, so there is only 2 roots.

2. A+2i and A-2i are the only two roots of the equation because
firstly Im(root) = 2 and the conjugate root theorem applies as the coefficients of the polynomial are real

3. Noting that sum of roots = -6, this gives the real part (A) as A=-3.

Therefore, roots are -3+2i and -3-2i
This is much neater and quicker.


Product of the roots then follows (3+2i)(3-2i) = 13.

Integrand's proof is virtually identical but a 'wee' bit clearer/elegant.

 

[InteGrand]

I have a question that I feel i worked out rather weirdly, so I was hoping there was an easier way to do these types of questions. I will post the question and my solution.

Question:

View attachment 32358

My solution:

http://m.imgur.com/Ej5ZhAt

My solution is correct by the way, just looking for other methods.

 

[InteGrand]

there is a simpler method
1. This polynomial is of degree 2, so there is only 2 roots.

2. k+2i and k-2i are the only two roots of the equation because
firstly Im(root) = 2 and the conjugate theorem applies as the coefficients of the polynomial are real

3. Using sum of roots = -6, gives the real part as follows:
2k=-6
k=-3

Therefore, roots are -3+2i and -3-2i
This is much neater and quicker

 

[calamebe]

Oh wow thanks a lot guys. That theorem slipped my mind, probably as it had the k at the end, I'll keep that in mind for later, thanks.

 

[dan964]

already fixed (at time of when I spotted your post)