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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.


  1. (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.

$$(x-(\frac{-1+5i}{2})(x-(-\frac{-1-5i}{2})$$ $$(x+\frac{1}{2} - \frac{5i}{2})(x+\frac{1}{2}+\frac{5i}{2})$$

$$x^2+x+\frac{13}{2}$$ $$2x^2+2x+13$$

$p=2,q=2, r=13$

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    $\begingroup$ @Mann - Good point and the product of two conjugates returns a real number (or a complex with no imaginary if you wish) $\endgroup$ – Juxhin May 12 '15 at 17:32
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    $\begingroup$ The beginning looks fine. There is an arithmetical error in calculating $b$. $\endgroup$ – André Nicolas May 12 '15 at 17:33
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    $\begingroup$ Woops, should be $-1+5i$ I believe! $\endgroup$ – Juxhin May 12 '15 at 17:36
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    $\begingroup$ They were minor things, your calculation had the right structure. $\endgroup$ – André Nicolas May 12 '15 at 17:43
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    $\begingroup$ I know the feels @Juxhin , I'd be having billion reputation as such. $\endgroup$ – Mann May 12 '15 at 17:46
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I will first note that $(2+3i)(1+i) = -1+5i$, so your value of $a+bi$ needs to be adjusted.

If you have a quadratic and you know its two roots $\alpha, \beta$, it must be equivalent (up to constant scaling) to $(x-\alpha)(x-\beta)$. Thus, your polynomial is (a scaled multiple of) $$ (x-a+bi)(x-a-bi) = x^2 - 2ax +a^2+b^2 $$

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  • $\begingroup$ I have adjusted that, silly mistake from my part. I managed to grasp your concept so far. I'm going to attempt to work it out with your method and see if I can match the final result $\endgroup$ – Juxhin May 12 '15 at 17:37
  • $\begingroup$ Thanks managed to get to the answer thanks to your hint $(x-\alpha)(x-\beta)$ :-) $\endgroup$ – Juxhin May 12 '15 at 17:52
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There are infinitely many solutions to this unless you specify the coefficients are relatively prime integers. In that case:

$$P(x)=(x+1-5i)(x+1+5i).$$

Now multiply it out.

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  • $\begingroup$ Thanks Tim, will be attempting both yours and Rolf's method out to see if I can get to the right result $\endgroup$ – Juxhin May 12 '15 at 17:38
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    $\begingroup$ Edited my answer to account for your correction of the value of $a+bi$. $\endgroup$ – Tim Raczkowski May 12 '15 at 17:41
  • $\begingroup$ Got the right answer! $\endgroup$ – Juxhin May 12 '15 at 17:47
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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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