3

Consider the following equation, where $z \in \mathbb{C}$, $i$ is the imaginary unit and $\overline{z}$ is the conjugate of $z$:

$$ z^2 + (1+i) \overline{z} + 4i = 0 $$

What is the method to deal with equations such as this?

I have tried various things: I tried substituting $z$ with $a+bi$, or $re^{i\theta}$, hoping I'd notice something. I thought I could somehow transform this into a quadratic equation, but I couldn't. Now I have no idea what to try. I'd appreciate ideas greatly.

Prasun Biswas
  • 6,334
bp99
  • 1,107

2 Answers2

5

$$a^2-b^2+2iab+(1+i)(a-ib)+4i=0\\ a^2-b^2+a+b+(2ab+a-b+4)i=0\\ \begin{cases}a^2-b^2+a+b=0\\ 2ab+a-b+4=0\end{cases}\\ \begin{cases}(a+b)(a-b+1)=0\\ 2ab+a-b+4=0\end{cases}\\ \begin{cases}a=-b\\ -2a^2+2a+4=0\end{cases}\vee\begin{cases}b=a+1\\ 2a^2+2a+3=0\end{cases}$$

An these can be solved for $a,b\in\Bbb R$ accordingly.

  • For future reference for people who might have the same problem as I did: I didn't think about the fact that $0$ is also a complex number, since it's $0 + 0i$. And then it's simple: Two complex numbers are equal IFF both their real and their imaginary coefficients are equal to one another. Thanks for your answer! – bp99 Sep 07 '17 at 21:05
4

If we conjugate given equation we get $$\overline{z} ^2+(1-i)z-4i=0$$

Express from starting equation $2\overline{z} = -(1-i)z^2-4-4i$

So we have: $$ \big[-(1-i)z^2-4-4i\big] ^2 + (4-4i)z-16i=0$$

Now solve it on $z$... but I belive G. Sassatelli solution is nicer.

After some calculation we get:

$$ z^4+4(1+i)^2z^2+2(1+i)z-8=0$$

I frequently use the fact that $(1\pm i)^2 = \pm 2i$.

Now let $w= 2(1+i)z$, then we get (notice that $w^4 = -64z^4$)

$$ w^4-64w^2-64w+512=0$$ which can be easly factorised as $$ (w-4)(w+8)(w^2+4w-16)=0$$

nonuser
  • 90,026
  • I'm not quite sure your method works. The equation turns out to have only two solutions, but yours has four. –  Sep 07 '17 at 18:33
  • As I can see yours has also 4 – nonuser Sep 07 '17 at 18:34
  • No, because $2a^2+2a+3$ has no real roots. –  Sep 07 '17 at 18:35
  • Ah, I didn't go into the details... – nonuser Sep 07 '17 at 18:35
  • They might be worth your attention, though, might they not? Since you are sketching a method for someone who does not know it, the results should at least be correct. –  Sep 07 '17 at 18:43
  • I'm working on it, but as I said, your way is safer... – nonuser Sep 07 '17 at 18:47
  • @G.Sassatelli Didn't realize that the question had been asked before when I posted my answer here, until it was pointed out that it's a duplicate. FWIW using the same "trick" $(1+i)^2=2i$ invoked in the answer above, but in a different way, avoids the problem of extraneous solutions. – dxiv Sep 13 '17 at 22:49