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The question is stated as following:

"First, solve the equation:

$ω^2=-11/4+15i$

and after, with the help of that, solve:

$z^2-(3-2i)z+(4-18i)=0$"

The problem for me lies in solving the system of equations for ω;

$Re:a^2-b^2=-11/4$ and $Im:2ab=15$

Where I eventually end up with the fourth degree equation: $4b^4+11b^2+15^2=0$ which I don't know how to solve, is this solveable or am I on the wrong track? I would very much appreciate some help, thank you for your time.

J. Dilla
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    Consider this equation as a quadratic in $b^2$. – Donald Splutterwit Sep 16 '18 at 19:59
  • What do you mean by this? @DonaldSplutterwit – J. Dilla Sep 16 '18 at 20:07
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    Let $x=b^2$ in your fourth power equation and you get a quadratic as Donald says. Solve it and then substitute back. – Mandelbrot Sep 16 '18 at 20:12
  • @J.Dilla: Can I suggest that maybe the problem suggests the easier approach of realizing that when you solve the quadratic equation for $z$, the value is found as $z=\frac{1}{2}(3-2i) \pm \frac{1}{2}\sqrt(-11+60i)$. This simplifies to $z=\frac{1}{2}(3-2i) \pm \sqrt(\frac{-11}{4}+15i)$. Then you can use the solutions you obtained for $\omega^2=\frac{-11}{4}+15i$ to simplify the solution for $z$. – Winter Soldier Sep 16 '18 at 21:17
  • @WinterSoldier The problem definetly does, it's the solutions for ω I'm having trouble solving, but I think that the hint that bruce gave me should give me the right answer – J. Dilla Sep 16 '18 at 21:28
  • @J.Dilla: Sorry, I missed that point. – Winter Soldier Sep 16 '18 at 21:39
  • Related: https://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number – Hans Lundmark Sep 17 '18 at 06:02

3 Answers3

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The standard method for square roots of complex numbers adds a third equation: the module of $\omega$, squared, is equal to the module of $-\frac{11}4+15i$: \begin{gather} a^2-b^2=-\frac{11}4,\qquad ab=\frac{15}2,\\[1ex] a^2+b^2=\sqrt{\frac{121}{16}+225}=\sqrt{\frac{3721}{16}}=\frac{61}{4}. \end{gather}

So we obtain a linear system for $a^2$ and $b^2$, whence $a$ and $b$, if you take into account that the second equation implies $a$ and $b$ have the same sign.

Bernard
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Note that $$|2\omega|^2=\left|-11+60\text{i}\right|=61\,.$$ We make an Ansatz that $2\omega$ is a Gaussian integer, namely, $2\omega=a+b\text{i}$ for some $a,b\in\mathbb{Z}$. It is not difficult to find integers $a$ and $b$ such that $$a^2+b^2=61\,.$$ Without loss of generality, you may assume that $a\geq 0$. (Well, since $61$ is a prime natural number congruent to $1$ modulo $4$, the pair $(a,b)$ is unique up to sign and permutation.)

If $a\geq0$, then there are four possibilities: $(a,b)=(5,\pm6)$ and $(a,b)=(6,\pm5)$. It is easy to see that $(a,b)=(5,6)$ works. Therefore, $2\omega= \pm(5+6\text{i})$, or $$\omega=\pm\left(\frac{5}{2}+3\text{i}\right)\,.$$

In general, if you have coprime integers $m$ and $n$ such that $m$ is odd and $m^2+n^2=k^2$ for some integer $k>0$, then from https://en.wikipedia.org/wiki/Pythagorean_triple, we have $$(m,n)=(a^2-b^2,2ab)$$ for some coprime $a,b\in\mathbb{Z}$ with $a\not\equiv b\pmod{2}$. That is, $$m+n\text{i}=(a+b\text{i})^2\,.$$ Note that $a^2+b^2=k$. Here are a few examples: $-3-4\text{i}=(1-2\text{i})^2$, $5+12\text{i}=(3+2\text{i})^2$, $15-8\text{i}=(4-\text{i})^2$, and $-7+24\text{i}=(3+4\text{i})^2$.

Batominovski
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As you have noted, equating the real parts gives $a^2 - b^2 = -\frac{11}{4}$. Now since the modulus of $\omega^2$ is the squared modulus of $\omega$ (think in polar form), $|\omega|^2 = a^2 + b^2 = \sqrt{(-11/4)^2 + 15^2} = \frac{61}{4}$.

Thus adding and subtracting, $a^2 = \frac{1}{2} \left(\frac{61}{4} + -\frac{11}{4} \right) = \frac{25}{4}$ and $b^2 = \frac{1}{2} \left(\frac{61}{4} - \frac{11}{4}\right) = 9$. But since the argument of $\omega^2$ is either in the second quadrant or the second quadrant + $360º$, the argument of $\omega$ is half that of $\omega^2$, which results in the first quadrant or the third quadrant. Hence $a+bi = \frac{5}{2} + 3i, -\frac{5}{2} - 3i$.

Toby Mak
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