<s>[INST] Finding real coefficients of equation given that $a+ib$ is a root Below is the question present in a past examination paper. I'll be giving my attempts and how I thought it through. Do feel free to point out any mistakes I make throughout my working even if unrelated to the question itself. 



  
*(a) Find $a+ ib$ such that $\frac{2+3i}{1-i} = a+ib$. If $a+ib$ is a root of the equation $px^2+qx+r=0$ where $p$ and $q$ are real numbers, find $p$,$q$ and $r$.
  


Before getting down to the technical side of the problem, I analyzed the question (somewhat). If the complex number $a+ib$ is a root of the given equation, then its conjugate $a-ib$ is also a root of the equation. However, since this is a quadratic equation, does the number of roots exceed the highest degree (in this case, 2)? If not, then are $a+ib$ and $a-ib$ the only roots of the equation?

Attempt to find $a+ib$
$$\frac{2+3i}{1-i} = a + ib$$
$$\frac{2+3i}{1-i} * \frac{1+i}{1+i} = a+ib$$
$$\frac{(2+3i)(1+i)}{2} = \frac{-1+5i}{2} = a+ib$$
$\therefore \frac{-1+5i}{2}$ is the first root and $\frac{-1-5i}{2}$ is the second root.
With these two roots, how would I proceed to obtain the coefficients of the above equation? I know it seems as though I haven't attempted this, I have however I cannot find any resources online that relate to this particular problem (somewhat identically atleast).
Method of solving (results matched with answersheet)
Since I got both roots $\frac{-1+5i}{2} = \alpha$ and $\frac{-1-5i}{2} = \beta$. I was able to structure my quadratic as $(x-\alpha)(x-\beta)$.
From there on all I did was subsitute, expand, simplify.
$$(x-(\frac{-1+5i}{2})(x-(-\frac{-1-5i}{2})$$
$$(x+\frac{1}{2} - \frac{5i}{2})(x+\frac{1}{2}+\frac{5i}{2})$$
$$x^2+x+\frac{13}{2}$$
$$2x^2+2x+13$$
$p=2,q=2, r=13$
 [/INST] First $px^2 + qx + r = 0$ and $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ have the same set of solutions. Note that we are sure that $p \neq 0$ (Why??) Think.
We have $\frac{q}{p} = 1$ and $\frac{r}{p} = \frac{1}{2}.$ We get then the equation $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ as $x^2 + x + \frac{1}{2} = 0,$ so one of several possibilities of $px^2 + qx + r = 0$ is $2x^2 + 2x + 1 = 0.$
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