I'm trying to solve this problem $$ z^2 + (\sqrt{3} + i)|z| \bar{z}^2 = 0 $$

So, I know $ |z^2| = |z|^2 = a^2 + b ^2 $ and $ \operatorname{Arg}(z^2) = 2 \operatorname{Arg} (z) - 2k \pi = 2 \arctan (\frac{b}{a} ) - 2 k\pi $ for a $ k \in \mathbb{Z} $. Regarding the other term, I know $ |(\sqrt{3} + i)|z| \bar{z}^2 | = |z|^3 |\sqrt{3} + i| = 2 |z|^3 = 2(a^2 + b^2)^{3/2} $ and because of de Moivre's theorem, I have $ \operatorname{Arg} [(\sqrt{3} + i ) |z|\bar{z}^2] = \frac{\pi}{6} + 2 \operatorname{Arg} (z) - 2Q\pi $.

Using all of this I can rewrite the equation as follows

$$\begin{align*} &|z|^2 \Bigl[ \cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg}(z) - 2k \pi)\Bigr]\\ &\qquad \mathop{+} 2|z|^3 \Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0 \end{align*} $$

Which, assuming $ z \neq 0 $, can be simplified as $$\begin{align*} &\cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg} (z) - 2k \pi) \\ &\qquad\mathop{+} 2 |z|\Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q \pi \right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0 \end{align*} $$

Now, from this I'm not sure how to go on. I tried a few things that got me nowhere like trying to solve $$ \cos (2 \operatorname{Arg}(z) - 2k \pi) = 2 |z| \cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) $$

I'm really lost here, I don't know how to keep going and I've looked for error but can't find them. Any help would be greatly appreciated.

  • $\begingroup$ Hint: write $z = re^{i\theta}$ (so that $|z| = r$). You'll eventually be able to solve for $re^{-4i\theta}$. $\endgroup$ – Zarrax Jun 29 '11 at 20:54
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    $\begingroup$ Note that $z=0$ is an obvious solution. $\endgroup$ – Mark Bennet Jun 29 '11 at 20:57
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    $\begingroup$ You have $z^2 = - (\sqrt{3} + i)|z| \bar{z}^2$ So that if $z \neq0$ then $\frac{z^2}{\bar{z}^2} = - (\sqrt{3} + i)|z|$ The modulus of the left hand side is 1, so that might help. $\endgroup$ – Mark Bennet Jun 29 '11 at 20:59

The relation is equivalent to $z^2=-(\sqrt{3}+i)|z|\overline{z}^2$. $z=0$ is a solution, so in the following $z \neq 0$. Take modulus of both sides and denote $r=|z|=|\overline{z}|$. Then $r^2=2r^3$, which means $r=\frac{1}{2}$.

The relations turns to $z^2+\frac{1}{2}(\sqrt{3}+i)\overline{z}^2=0$. Multiply by $z^2$ and get $z^4+\frac{1}{2}(\sqrt{3}+i)\frac{1}{16}=0$. Write it in trigonometric form $$ z^4=\frac{1}{16}\left(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)=\frac{1}{16}(\cos \frac{7\pi}{6}+\sin \frac{7\pi}{6})$$.

From here on it is just the extraction of complex roots.

[edit] I did not answer your question, as to how to continue your calculations, but I can say from experience that in most complex numbers problems the substitution $z=a+bi$ gets you in more troubles in the end, than working with the properties of complex conjugate, modulus and trigonometric form. You can see that in my solution no great computational problems were encountered.

  • $\begingroup$ I had something similar, but deleted it for the reason that it does not really answer the question asked! $\endgroup$ – Aryabhata Jun 29 '11 at 21:25
  • $\begingroup$ I just wanted to illustrate how to approach these problems in general. His computations are almost impossible to be finalized in a solid solution. $\endgroup$ – Beni Bogosel Jun 29 '11 at 21:33
  • $\begingroup$ @Aryabhata You're right, it doesn't exactly give me the exact solution but it does simplify the problem to a pretty simple one in a very elegant way. Thanks for taking the time to go through this, to you and @Beni $\endgroup$ – Bananas Jun 29 '11 at 21:36
  • $\begingroup$ I was of a mind to multiply by $z^2$ from the get-go, with the left-hand side transforming to $z^4 + 2|z|^5 \exp(i \pi/6)$, then to $|z|^4 \exp(4 i \arg(z)) + 2|z|^5 \exp(i\pi/6)$, and then to $|z|^4 \left( \exp(4 i \arg(z)) + 2|z| \exp(i\pi/6) \right)$. From there, we have either that $z=0$ or that $|z| = 1/2$ and $4 \arg(z) = 7\pi/6 \mod 2\pi$. We then have to check which these work in the original equation. $\endgroup$ – Blue Jun 29 '11 at 23:17

Here is an alternative to solving it using polar form. Let $z=a+bi$, so that $\bar{z}=a-bi$ and $|z|=\sqrt{a^2+b^2}$. Then you want to solve $$(a+bi)^2+(\sqrt{3}+i)\sqrt{a^2+b^2}(a-bi)^2=0,$$ which expands to $$(a^2-b^2)+2abi+(\sqrt{3}+i)\sqrt{a^2+b^2}\left((a^2-b^2)-2abi\right)=0$$ Thus, we need both the real part and the imaginary part of the left side to be 0, i.e. $$(a^2-b^2)+\sqrt{a^2+b^2}\left(\sqrt{3}\cdot (a^2-b^2)+2ab\right)=0$$ and $$2ab+\sqrt{a^2+b^2}\left(-2ab\sqrt{3}+(a^2-b^2)\right)=0.$$ It should be possible to solve these equations by simple manipulations, though I haven't worked it out myself yet.

  • $\begingroup$ That route was the first thing I tried, but I though finding a solution to this new equation was too hard for it to be the "correct". I even tried clearing $ a^2 - b^2 $ from the first one, but after a while I left it since I got nowhere. If you know of a relatively simple way of solving this I'd love to hear it, since I can't think of any. $\endgroup$ – Bananas Jun 29 '11 at 21:46

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