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Could anyone help me to solve two equations with complex numbers? I would like to know if there is a way to solve them avoiding the usual substitution $z=a+ib$ because calculations are not very easy

$1)$$z^3\bar{z}+3z^2-4=0$

I tried in this way

$z^2(z\bar{z}+3)=4$

$z^2(|z^2|+3)=4$

But I'm stuck at this point

$2)$$ \begin{cases} |z^2+1|=1 \\ 2Re(z)=|z^2| \end{cases}$

Here I tried to substitute $\omega=z^2$ but it doesn't work

Thanks a lot in advice

Gianolepo
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2 Answers2

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1) Your equation $z^2(z\bar{z}+3)=4$ brings us close to the end. Let $|z^2|=c$, and take the norm of both sides. Then $c(c+3)=4$, giving $c=1$.

2) This one is a little messier. There is the obvious solution $z=0$. We look for non-zero solutions. Expanding the first equation tells us that $$z^2\bar{z}^2+(z^2+\bar{z}^2)=0.$$ From the second equation, $z^2\bar{z}^2=4(\text{Re}(z))^2$. Letting $z=re^{i\theta}$ and cancelling the $r$ we get $$4\cos^2\theta+2\cos(2\theta)=0.$$ The rest is straightforward. Use $\cos(2\theta)=2\cos^2\theta-1$.

André Nicolas
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  • Thanks! But why did you substitute also $z^2$ with $c$ ? Is in this case $|z^2(|z^2|+3)|=|z^2|(|z^2|+3)$ ? – Gianolepo Nov 21 '15 at 16:54
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    Yes, the norm of a product is the product of the norms. – André Nicolas Nov 21 '15 at 16:58
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    So we end up with the equation $z^2=1$. Not hard! – André Nicolas Nov 21 '15 at 17:15
  • Not at all! Thanks again! Do you have any suggestion for the problem $2)$ ? – Gianolepo Nov 21 '15 at 17:17
  • @FrancescoCaruso: I have written out most of the solution to 2). One could have waited another step or so before introducing the cosine, so that instead of using trig identity we use properties of exponentials. – André Nicolas Nov 21 '15 at 18:52
  • Thank you I tried to do it but I get something different: $(z^2+1)\bar{(z^2+1)} -1=(z^2+1)(\bar{z^2}+1) -1=z^2\bar{z^2}+z^2+\bar{z^2}+1-1=z^2\bar{z^2}+z^2+\bar{z^2}$ where does that $2(z^2+\bar{z^2})$ comes from? – Gianolepo Nov 21 '15 at 19:24
  • It comes from a mistake, one too many $2$'s! Fixed. . – André Nicolas Nov 21 '15 at 19:27
  • Ok! One last thing, in the second equation of problem $2)$ i put $z=\rho e^{i\theta}$ and I get $\rho^{4} =4(\rho cos\theta)^{2}$ from which $\rho^{4}=4\rho^{2}$ and $cos^{2}\theta=0$ I don't get the $\rho$ canceled and I don't have the $2cos(2\theta)$ . Is this wrong? – Gianolepo Nov 23 '15 at 08:50
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    @FrancescoCaruso: I may be repeating a mistake, but it still cancels, giving $\cos\theta=\pm \frac{1}{2}$. But the second equation forces positive, and forces norm equal to $1$. So we get the two non-real cube roots of unity. Now in case you are concerned, substitute in the first equation. They work. – André Nicolas Nov 23 '15 at 16:55
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    Oh, I see what your problem is, you are looking at the wrong equation. It is from $4(\text{Re}(z))^2+z^2+\bar{z}^2=0$ that I get the cancellation of $r^2$, and then trig equation. – André Nicolas Nov 23 '15 at 17:24
  • You're right I was looking at the wrong equation! Now it's clear! Thanks again for your help! – Gianolepo Nov 23 '15 at 17:43
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Actually, what you have done is pretty clever/lucky in that you know $(|z^2| + 3)$ is real so $z^2(|z^2| + 3) = 4$ means $z^2$ is real. So $z^2 = \pm |z^2|$. Set $w = z^2$ and you have either:

$w \ge 0; w^2 + 3w - 4 = 0 \implies (x + 4)(x - 1) \implies w = 1 \implies z = \pm 1$

$w < 0; w^2 - 3w - 4 = 0 \implies (x - 4)(x + 1) \implies w = -1 \implies z = \pm i$

So $z = \pm i, \pm 1$.

fleablood
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