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Let $a + bi$ be an algebraic number. Then there is polynomial which coefficients are rational number and one of root is $a+bi$.

I think.. $$x = a + bi$$ we can subtract $c_1$ (which is rational number) from both sides. $$x-c_1=a-c_1+bi$$ and we can power both side. $$(x-c_1)^n=(a-c_1+bi)^n$$

and repeat we can get polynomial which coefficients are rational number and one of root is $a+bi$.

Therefore, I think there is no 'Algebraic number' which cannot be displayed with arithmetic operations and roots.

Matt Samuel
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1 Answers1

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The algebraic numbers are divided into the explicit algebraic numbers and the implicit algebraic numbers.

The explicit algebraic numbers can be presented explicitly from the rational complex numbers by the arithmetic operations (addition, subtraction, multiplication, division, raising to integer powers, and $n$-th roots where $n$ is an integer). They are called solutions in radicals.

All solutions of algebraic equations of degree $\leq$ 4 are solutions in radicals. There are equations of degree $\ge$ 5 that have solutions that are not solutions in radicals. This is stated by Abel–Ruffini theorem.

The algebraic equation $x^5-x+1$ for example doesn't have solutions in radicals.

IV_
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