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?
 A: 

*

*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.

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