moving i's outside of surds in radical expressions

Question

Given an algebraic number expressed in radicals that has an i inside a surd, does an alternative representation of the same value that is still expressed in algebraic terms but with the i not inside a surd always exist? For example: $$\sqrt{1+i}=\sqrt[4]{2}\sqrt{2+\sqrt{2}}/2+i\sqrt[4]{2}\sqrt{2-\sqrt{2}}/2$$ If such a representation does always exist, is there a general method for finding it?

Edit: I am interested in cases where the surd is an arbitrary nth root, not just a square root as in the example.

Possible duplicate of How do I get the square root of a complex number? — José Carlos Santos, Nov 21 '17 at 09:51
@JoséCarlosSantos : This isn’t even close to being a duplicate of the question you cite — MPW, Nov 21 '17 at 09:55
@MPW This answer to that question contains exactly what the OP wants. — José Carlos Santos, Nov 21 '17 at 09:57
@JoséCarlosSantos It does provide exactly what I want when the surd is a square root. I'd like to know about arbitrary nth roots too, though. If it is in general necessary to solve a degree n polynomial to do it, then presumably it isn't possible in general after all. — Alex Kindel, Nov 21 '17 at 10:01
@AlexKindel Because of your example, I thought that you were interested in square roots only. I shall withdraw my closing vote. — José Carlos Santos, Nov 21 '17 at 10:02

score 1 · Answer 1 · answered Nov 21 '17 at 13:08

1

If $z=\sqrt[n]w$ then, in general you are trying to solve $z^n=w=\rho(\cos\phi + i\sin\phi)$ say, i.e. express $w$ in polar form.

The $n$ roots are then

$$ z_k=\rho^\frac{1}{n} \left( \cos\left(\frac{\phi}{n} + k\frac{2\pi}{n}\right) +i\sin\left(\frac{\phi}{n} + k\frac{2\pi}{n}\right) \right)\\ k=0,\ldots,n-1 $$ with the principal root being given by $k=0$.

In your case $w=1+i = \sqrt2(\cos\pi/4+i\sin\pi/4)$

answered Nov 21 '17 at 13:08

PM.

5,249

Whether that sine and cosine are expressible in radicals depends on whether their arguments are π times a rational for which the denominator of the reduced form is a power of 2 or a power of 2 times a product of distinct Fermet primes. Suppose z=a+bi with a and b real. Then we have ϕ=arctan2(b,a). Is it obvious, then, whether a and b expressible in real radicals results in a ϕ such that ϕ/n+k2π/n satisfies the required criterium? It does look very unlikely that any values will do. – Alex Kindel Nov 21 '17 at 17:22
@PM.: How does this even address the question, let alone answer it? – Alex M. Nov 21 '17 at 21:47
@AlexM. It shows how to obtain the roots of a complex number with i not inside a surd. This is what was asked, or so i thought. – PM. Nov 21 '17 at 22:35
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@PM. That is part of the question, but the other part is whether the expression for the roots with i not inside a surd can be written in terms of radicals. In the given example, it can because we know how to write cos(π/4) and sin(π/4) in terms of radicals, but it's not clear to me in what cases it is or isn't possible, mostly because I don't have a great feel for how arctan, and hence ϕ, behaves. – Alex Kindel Nov 21 '17 at 23:22

score 0 · Answer 2 · answered Dec 04 '17 at 13:32

For $z = a + bi$, $$\sqrt[n]{z} \\= (|z|e^{i * atan2(a, b)})^{1 / n} \\= |z|^{1 / n} * e^{i * atan2(a, b) / n} \\= |z|^{1 / n} * (\cos(\frac{atan2(a, b)}{n}) + i\sin(\frac{atan2(a, b)}{n})$$ The question is equivalent to whether $\cos(\frac{atan2(a, b)}{n})$ and $\sin(\frac{atan2(a, b)}{n})$ are expressible in real radicals. In cases where $\frac{atan2(a,b)}{n}$ is a rational multiple of $\pi$, this is true if and only if the denominator of the reduced form of the rational multiple is a power of 2 or a power of 2 times a product of distinct Fermat primes. So an example that establishes that the answer to the question is no can be obtained by choosing $a$ and $b$ such that $atan2(a,b)$ is a rational multiple of $\pi$, say, $z = 1 + i$ so that $atan2(a,b)=\pi/4$, and choosing a value for $n$ such that the resulting rational multiple does not satisfy the criterium described above, like $n=7$.

It would still be interesting to have a test for whether it is possible to denest $i$ in a given radical expression. An algorithm exists for finding a radical expression of minimal nesting depth with the same value as an arbitrary nested radical. Depending on how close to unique the result can be shown to be, this algorithm might suffice to answer the question.

moving i's outside of surds in radical expressions

2 Answers2