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?
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.
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 $p^2(x)+q^2(x)$.
Multinomial Combiniations of Algebraic Integers
Given a set of monic polynomials defining a set of algebraic integers, how can we derive a polynomial for a multinomial combination of those algebraic integers?
For example, given $\alpha^2-3=0$ and $\beta^3-2=0$, how do we derive a polynomial for $\alpha+\beta$?
First, create the basis for all multinomial combinations of $\alpha$ and $\beta$: $$ B=\begin{bmatrix} 1\\ \alpha\\ \beta\\ \alpha\beta\\ \beta^2\\ \alpha\beta^2 \end{bmatrix} $$ Next compute $(\alpha+\beta)B$: $$ \begin{align} (\alpha+\beta)B &=(\alpha+\beta) \begin{bmatrix} 1\\ \alpha\\ \beta\\ \alpha\beta\\ \beta^2\\ \alpha\beta^2 \end{bmatrix} =\begin{bmatrix} \alpha+\beta\\ \alpha^2+\alpha\beta\\ \alpha\beta+\beta^2\\ \alpha^2\beta+\alpha\beta^2\\ \alpha\beta^2+\beta^3\\ \alpha^2\beta^2+\alpha\beta^3 \end{bmatrix} =\begin{bmatrix} \alpha+\beta\\ 3+\alpha\beta\\ \alpha\beta+\beta^2\\ 3\beta+\alpha\beta^2\\ \alpha\beta^2+2\\ 3\beta^2+2\alpha \end{bmatrix}\\[6pt] &=\begin{bmatrix} 0&1&1&0&0&0\\ 3&0&0&1&0&0\\ 0&0&0&1&1&0\\ 0&0&3&0&0&1\\ 2&0&0&0&0&1\\ 0&2&0&0&3&0\\ \end{bmatrix} \begin{bmatrix} 1\\ \alpha\\ \beta\\ \alpha\beta\\ \beta^2\\ \alpha\beta^2 \end{bmatrix} =MB \end{align} $$ $M$ is an integer matrix since the polynomials for $\alpha$ and $\beta$ are monic.
Since multiplication by a scalar commutes with matrix multiplication, induction gives $$ (\alpha+\beta)^kB=M^kB $$ SInce $M$ satisfies its own characteristic polynomial, $$ 0=P(M)B=P(\alpha+\beta)B $$ therefore, $\alpha+\beta$ also satisfies the characteristic polynomial of $M$: $$ P(x)=x^6-9x^4-4x^3+27x^2-36x-23 $$ Since the characteristic polynomial, $P(x)=\det(M-Ix)$, for an integer matrix is monic, $\alpha+\beta$ is also an algebraic integer.
Application to the Question
The example above should be general enough to show how to apply the process to any set of monic polynomials, a cubic and quadratic were used to reduce the footprint of the example.
To show that $5^{1/5}-3i$ is an algebraic integer, apply the program above to $\alpha^5-5=0$ and $\beta^2+9=0$ to produce a monic polynomial for $\alpha-\beta$: $$ x^{10}+45x^8+810x^6-10x^5+7290x^4+900x^3+32805x^2-4050x+59074 $$
Afterword
I had thought I had written something along these lines before. I finally came across this answer that I wrote a while back. Hopefully, between this post and that one, the ideas are sufficiently clear.
