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It's a classic problem in many textbooks: Solve in complex numbers the equation $$z^5=\bar{z} \tag{1}$$ The solution is a as follows: We apply the modulus, and obtain that $|z|=0$, so $z=0$, which satisfies and second case $|z|=1$,so $\bar{z}=\frac{1}{z}$ and the equation now has the form $$z^6=1 \tag{2} $$ whose solutions are the 6-th roots of unity. So final solution is $$z=0,\ U_6.$$

Ok, now my main concern: shouldn't we "check" the final solutions?

I mean, we applied the modulus, etc, found the possible value of modulus, then using the equation again and derived only a CONSEQUENCE of the original equation, so all the solutions of original equation are also solutions of $z^6=1$ but not necessarily the other way around.

I mean, the checking is very simple, but what surprised me is that no solution I read does this check.

So I began to ask myself if there's something I don't see? Is that something obvious guaranteeing that equations (1) and (2) indeed have exactly the same solution set, regarding that, in the process of solution, equation (2) comes only a s a consequence of (1)? I mean, could I find another problem solved in the same manner, such that at the end it gives extraneous roots?

So my main question is about the equivalence of equations. What bothers me is that in the 4-5 textbooks I found this solutions, they check $z=0$, but don't bother to check $U_6$.

amarius8312
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1 Answers1

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Your question is whether you have introduced extraneous solutions in the process. The answer is negative.

Extraneous solutions are introduced when a "step can not be reversed". E.g. if you square both sides of $\sqrt{x}=-5$, you get $x=25$, but the step can not be reversed. I.e. $x=25$ does not imply that $\sqrt{x}=-5$.

In your case, under the assumption that $|z|=1$, we get the equivalence: $z$ satisfies $(1)$ if and only if $z$ is a $6^{th}$ root of unity. Since this is an equivalence, no extraneous solutions have been introduced.

Ioannis Souldatos
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  • "under the assumption that |z|=1, we get the equivalence: z satisfies (1) if and only if z is a 6th root of unity". – amarius8312 Mar 01 '16 at 16:50
  • @amarius8312: Are you asking a question about the quoted phrase? – Ioannis Souldatos Mar 01 '16 at 16:54
  • Yes, excuse me.This part seems a bit tricky to me! What if we "worked in the assumption that |z|=2"?(discard for a moment that this can't take place". We would obtain that "in the assumption that |z|=2, z satisfies (1) if and only if z is a 6'th root of 4. And thus concluding that the solutions are the 6-th roots of 4 which is of course very wrong. So?? When we work based on an assumption, aren't we supposed, at the end, to at least check that assumption, if not check the original form of equation? So i still think that equivalence "based on an assumption" is tricky, and we should check answer – amarius8312 Mar 01 '16 at 16:56
  • @amarius8312: Assume $|z|=1$. Then $z$ satisfies $(1)$ if and only if it satisfies $(2)$. This is bi-directional statement. You have proved that $z$ satisfies $(1)$ implies that $z$ satisfies $(2)$. My comment is that your steps are reversible. I.e. if $z$ satisfies $(2)$, $z^6=1$, $z^5=1/z$, since $z$ is non-zero, $z^5=\bar{z}$, since $|z|=1$. – Ioannis Souldatos Mar 01 '16 at 17:02
  • @amarius8312: The situation $|z|=2$ does not fit in this example, but say you had $z^5=2^4\bar{z}$. Then you get that either $|z|=0$ or $|z|=2$. Under the assumption $|z|=2$, you can prove that $z^5=2^4\bar{z}$ if and only if $z$ is a $6^{th}$ root of $2^6$. – Ioannis Souldatos Mar 01 '16 at 17:06
  • Yes, but we worked "under an assumption". And what if that assumption was false/not attainable at all ,i mean there were no solutions veryfinf the assumption, as in my example with |z|=2? My initial approach was: $z^6=1$, we take the modulus, we get $|z|=1$, and then $z^5=\frac{1}{z}$, $z^5=\bar{z}$$, so we proved the other direction without "working under an assumption" which may be problematical in general. – amarius8312 Mar 01 '16 at 17:24
  • @amarius8312: The assumption $|z|=1$ was used because this is what you got by taking the modulus. If you were to assume $|z|=2$ by your modulus argument, there is no solution that satisfies $|z|=2$. Also, you are right that the other direction does not need any assumption, i.e. (2) implies (1) plus $|z|=1$. – Ioannis Souldatos Mar 01 '16 at 17:34