3
$\begingroup$

I am studying the book Complex Variables with Applications written by Herb Silverman. In this book, problem number 8 in Question 1.7 is as in the following.

Is $5^{1/5} - 3\cdot i$ algebraic? (i.e, Is the $5$th-root of $5$ minus $3\cdot i$ algebraic?)

Can you help me to solve this?

$\endgroup$
  • $\begingroup$ You should edit you question and include what your thoughts are so far, what you've tried etc. Also, algebraic over what? Algebraic over $\mathbb{Z}[x]$? $\endgroup$ – cansomeonehelpmeout Mar 4 '18 at 12:05
  • $\begingroup$ "Algebraic" by itself means "algebraic over the rationals". $\endgroup$ – GEdgar Mar 4 '18 at 12:42
  • $\begingroup$ Would that by any chance be this MIT Open Courseware? ocw.mit.edu/courses/mathematics/… $\endgroup$ – Robert Soupe Mar 4 '18 at 17:32
2
$\begingroup$
  • The number $5^{1/5}$ is algebraic since it is a zero of $z^5-5$.

  • The number $-3i$ is algebraic since it is a zero of $z^2+9$.

  • The sum of algebraic numbers is algebraic. See for instance this MSE post.

We conclude: $5^{1/5}-3i$ is algebraic.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is a nice, and simple solution. Well done! $\endgroup$ – Toby Mak Mar 4 '18 at 13:34
  • $\begingroup$ @TobyMak: Thanks. :-) $\endgroup$ – Markus Scheuer Mar 4 '18 at 13:34
2
$\begingroup$

Of course $\root 5 \of 5 - 3i$ is an algebraic number, and more than that, it is an algebraic integer. $\root 5 \of 5$ is an algebraic integer of degree 5 from the field of algebraic numbers $\mathbb Q(\root 5 \of 5)$, and $-3i$ is an algebraic integer from $\mathbb Z[i]$.

So then what field of algebraic numbers contains both $\root 5 \of 5$ and $-3i$? This goes beyond your question, but you might run into it later, if not in Silverman's book, in another book. $\mathbb Q(\root 5 \of 5, i)$ is a ring of algebraic degree 10, and the minimal polynomial of $\root 5 \of 5 - 3i$ is $x^{10} + 45x^8 +$ $\ldots +$ $32805x^2 - 4050x + 59074$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Let $a=\sqrt[5]5-3i$. Then\begin{align}a=\sqrt[5]5-3i\iff&a+3i=\sqrt[5]5\\\iff&(a+3i)^5=5\\\iff&a^5+15ia^4-90a^3-270ia^2+405a-5+243i=0.\end{align}So, if you define $p(x)=x^5-90x^3+405x-5$ and $q(x)=15x^4-270x^2+243$, $a$ is a root of $p(x)+q(x)i$. That is, $p(a)=-q(a)i$. But then $p^2(a)=-q^2(a)$. So, $a$ is a root of $p^2(x)+q^2(x)$.


This answer was completed with the help of the comments of the user Michael.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @cansomeonehelpmeout My mistake. Is it clear now? $\endgroup$ – José Carlos Santos Mar 4 '18 at 12:11
  • $\begingroup$ I think you have to be tricksier than that. From line 2 to 3 it seems that (a-3i)^5=5, but a-3i = 5^(1/5)-6i, whose fifth power isn't 5. $\endgroup$ – Michael Mar 4 '18 at 12:13
  • $\begingroup$ It seems $(a+3i)^2=(a+3i)(a-3i)$ $\endgroup$ – Joe Mar 4 '18 at 12:14
  • $\begingroup$ If $(a+3i)^5=5$, then $\overline{(a+3i)^5}=\overline5$. But $\overline{(a+3i)^5}=(\overline{a+3i})^5=(a-3i)^5$ and $\overline5=5$. Therefore $\bigl((a+3i)(a-3i)\bigr)^5=25$. $\endgroup$ – José Carlos Santos Mar 4 '18 at 12:19
  • $\begingroup$ @TobyMak Of course they have different roots. Why do you think I've put a $\implies$ sign there instead of a $\iff$? $\endgroup$ – José Carlos Santos Mar 4 '18 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.