Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have here a complex equation:

$$z^2 - (7+j)z + 24 +j7 = 0$$

How do we get the roots of this equation? I started using the quadratic formula $-b \pm \sqrt{ b^2-4ac}\over 2$, but it yielded too much complexity on it. Is there any way to directly attack this? Thanks.

share|cite|improve this question
What is $j$ (imaginary unit?). How can the solution be too complex (complicated?) – Dirk Jan 25 '12 at 13:40
yes j is imaginary unit – WantIt Jan 25 '12 at 21:37
up vote 4 down vote accepted

One can complete the square, that is, write $z^2-(7+j)z$ as the beginning of the expansion of $$ \left(z-\frac12(7+j)\right)^2. $$ This yields $$ z^2-(7+j)z+24+7j=\left(z-\tfrac12(7+j)\right)^2-u, $$ with $$ u=\tfrac14(7+j)^2-24-7j. $$ But $u=v^2$ for some complex number $v$, hence the equation to solve is equivalent to $$ \left(z-\tfrac12(7+j)\right)^2-v^2=0, $$ that is, $$ \left(z-\tfrac12(7+j)-v\right)\cdot\left(z-\tfrac12(7+j)+v\right)=0, $$ which yields the two solutions $$ z=\tfrac12(7+j)\pm v. $$ It remains to compute $v$...

share|cite|improve this answer
I will delete my answer because it does not add anything relevant to yours. – Américo Tavares Jan 25 '12 at 19:35

Although we do find the roots, the following is mainly a spoof of school algebra.

In school algebra, students are expected to solve very special equations of the form $ax^2+bx +c=0$, where $a$, $b$, and $c$ are integers, by factoring. The Quadratic Formula, and even the Rational Roots Theorem, are withheld from them, as that would make the problem too simple.

The process they are taught involves factoring $a$ and $c$, and fiddling a bit to try to produce $-b$. They are only given quadratics that yield to this process.

Let's play that game with our equation, to see whether we are dealing with a variant of a school problem. So we factor $24+7i$ in the Gaussian integers. Note that $(24+7i)(24-7i)=625$. If you have done some computations with Gaussian integers, you will see that the Gaussian primes involved in the factorization are $2\pm i$ (and associates, but we needn't worry about these). Also, since $5$ does not divide $24+7i$, we know that $24+7i$ must be an associate of $(2\pm i)^4$. Pretty quickly we find that $24+7i=-i(2+i)^4$.

Now let's find two Gaussian integers whose product is $-i(2+i)^4$ and whose sum is $7+i$. Note that $(2+i)^2=3+4i$ and $-i(2+i)^2=4-3i$. Their sum is $7+i$, so we have found the roots.

share|cite|improve this answer
in your 4th paragraph, you write "Also, since $24+7i$ does not divide $24+7i$...". I assume this is a typo. – Mark Beadles Jan 25 '12 at 17:40
@Mark Beadles: Thanks very much for careful reading! Yes, I meant since $5$ does not divide $24+7i$. That rules out $2+i$ and $2-i$ both appearing in the prime factorization. – André Nicolas Jan 25 '12 at 17:52

Note that the constant term $(24 + 7j)$ is one-half of the square of the $7 + j$ coefficient. This suggests writing $z = (7 + j)w$, and the equation becomes $$(48 + 14j) w^2 - (48 + 14j)w + (24 + 7j) = 0$$ Divide through by $(24 + 7j)$ and you get $$2w^2 - 2w + 1 = 0$$ By the quadratic formula this has roots ${1 \over 2} \pm {j \over 2}$. So the roots of the original equation are $(7 + j) ({1 \over 2} \pm {j \over 2})$, or in other words $3 + 4j$ and $4 - 3j$.

share|cite|improve this answer

Let's denote $z$ as : $z=a+jb$ , then we have :

$a^2-b^2+2abj-(7+j)(a+bj)+24+7j=0 \Rightarrow$

$\Rightarrow a^2-b^2+2abj - (7a+7bj+aj-b)+24+7j=0$

So , you have to solve following system of equations :

$\begin{cases} a^2-b^2-7a+b+24=0 \\ 2ab-7b-a+7=0\\ \end{cases}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.