# Is $5^{1/5} - 3\cdot i$ algebraic?

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?

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

• 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.

• This is a nice, and simple solution. Well done! – Toby Mak Mar 4 '18 at 13:34
• @TobyMak: Thanks. :-) – Markus Scheuer Mar 4 '18 at 13:34

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 $x^{10} + 45x^8 +$ $\ldots +$ $32805x^2 - 4050x + 59074$.

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.

• @cansomeonehelpmeout My mistake. Is it clear now? – José Carlos Santos Mar 4 '18 at 12:11
• 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. – Michael Mar 4 '18 at 12:13
• It seems $(a+3i)^2=(a+3i)(a-3i)$ – Joe Mar 4 '18 at 12:14
• 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 $\bigl((a+3i)(a-3i)\bigr)^5=25$. – José Carlos Santos Mar 4 '18 at 12:19
• @TobyMak Of course they have different roots. Why do you think I've put a $\implies$ sign there instead of a $\iff$? – José Carlos Santos Mar 4 '18 at 12:20