# Solutions to $\sqrt[4]{1} + \sqrt[4]{16}$ in Complex Numbers

In real numbers, $$\sqrt[4]{1} + \sqrt[4]{16}$$ = 3. In complex numbers we get more solutions.

a) Determine how many solutions there are.

b) Determine their absolute values.

According to the results, there are 16 distinct solutions to the sum. However, it also says that the absolute values of the solutions are {$$1; \sqrt{5}; 3$$}.

Does anyone know how to solve it?

Judging by the number of absolute values, I would say that there is a typo in the results and it should be 6 instead of 16.

Nonetheless, I have no idea how to get to those values. If it's too easy, just give me hint.

• Nope, there are $16$ solutions. There are four values for $\sqrt[4]{1}$ and four values for $\sqrt[4]{16}$. Commented Jan 4, 2017 at 15:36
• In general the number of complex n$^{\text{th}}$ roots of a number is n. Commented Jan 4, 2017 at 15:38
• All of the fourth roots of 1 are found through $z^4 - 1 = 0$ which, when factored via difference of squares, is very easy to solve. Commented Jan 4, 2017 at 15:40
• @Guru: to be exact, except x=0. Commented Jan 4, 2017 at 15:43

For a visual answer, I've shown the four roots of $1$ added to each of the four roots of $16$ in the complex plane, making a nice grid of $16$ points, characterized by $|Re(z)|+|Im(z)|=(1\text{ or }3)$. As you can see, there are four solutions with $|z|=1$, four with $|z|=3$ and eight off the axes lying on the circle with $|z|=\sqrt 5$:

Notice $\sqrt[4]{1}$ can take four values $\{1,-1,i,-i\}$. And the same with $\sqrt[4]{16}$, which can take $\{2,-2,2i,-2i\}$. So now you can form, how many sums choosing one element from each set?

• "how many sums" => "how many different sums". Commented Jan 4, 2017 at 15:46

16 is correct. Recall that the absolute value of a complex number is its magnitude - infinitely many share each nonzero magnitude. For example, $1$, $i$, $-1$, and $-i$ all have magnitude $1$.

$1$ has four fourth roots: $1, -1, i, -i$. $16$ also has four fourth roots; try to work them out. Each root of $1$ pairs with a root of $16$, for $4 \cdot 4$ total pairs.

Ignore the information about the magnitudes for now; just use it to check your work.

• You should demonstrate that addition of any pair is different to the addition of the others. Commented Jan 4, 2017 at 15:44
• That was much much easier than I thought. The problem was that, for some reason, I couldn't visualise it correctly despite knowing the theory behind it. Thank you! Commented Jan 4, 2017 at 15:54