# All Solutions for $(-256)^{\frac{1}{4}}$ and $1^{\frac{1}{5}}$ Imaginary Roots?

This is a question about the imaginary roots of the two equations $$(-256)^{\frac{1}{4}} \qquad\text{and}\qquad 1^{\frac{1}{5}}.$$ For the first one I've worked out that 2 of the solutions are $4\sqrt{\mathrm{i}}$ and $-4\sqrt{\mathrm{i}}$; but I thought the other two solutions would lie on the real axis and be $+4$ and $-4$. These were incorrect. What would the other two solutions be and why?

For the second one I found one of the solutions, which is $1$. There are another 4 solutions that I have no clue how to find.

I've found the solutions for (2): $1$, $\mathrm{e}^{\frac{2}{5}\pi\mathrm{i}}$, $\mathrm{e}^{\frac{2}{5}\pi\mathrm{i}}$, $\mathrm{e}^{\frac{4}{5}\pi\mathrm{i}}$, $\mathrm{e}^{-\frac{2}{5}\pi\mathrm{i}}$, $\mathrm{e}^{-\frac{4}{5}\pi\mathrm{i}}$. The third one is apparently incorrect. Why is this?

• $\pm4$ are not roots of $-256$, since $(\pm4)^4=256\neq-256$. For the second question, there should be four other roots, not five. – user170231 Dec 29 '15 at 19:25
• For the second question, you can try to write the roots as $z = re^{i\theta}$. – Watson Dec 29 '15 at 19:26
• Thanks, any clue what the other 2 roots are for the first one and how to get them? – Teyash Arjun Dec 29 '15 at 19:26

The $n$-th roots of a complex number $z$ are:

$$\sqrt[n]{r}\left(\cos\left(\dfrac{\alpha + 2\pi k}{n}\right) + i\sin\left(\dfrac{\alpha + 2\pi k}{n}\right)\right)$$

Where $k = 0,\ldots,n-1$. And $r = \Vert z\Vert$ and $\alpha = \arctan \frac{\Im (z)}{\Re (z)} +\pi\cdot s(z)$

Where $s(z) = 1$ if $\Im(z) < 0$ or if $\Im(z) = 0$ and $\Re(z) < 0$ and $s(z) = 0$ otherwise.

Also $\sqrt[n]{r}$ denotes the non-negative real valued $n$-th root of $r$.

• Just to help me refresh my memory: $\sqrt[n]r$ is the real-valued $n$th root of $r$, correct? – user170231 Dec 29 '15 at 19:31
• @user170231 Quite right. – Darth Geek Dec 29 '15 at 19:32
• I found the roots for number 2. The third answer is incorrect(Answer in question) Why is this @DarthGeek – Teyash Arjun Dec 29 '15 at 19:52

Write $-256 = 4^4e^{i(2n-1)\pi}$ for integer-valued $n$. Then, we have

\begin{align} \left(-256\right)^{1/4}&=\left(4^4e^{i(2n-1)\pi}\right)^{1/4}\\\\ &=4e^{i(2n-1)\pi/4} \end{align}

for $n=-1,0,1,2$. Then, the four solutions are $4e^{\pm i\pi/4}$ and $4e^{\pm i\pi/4}$.

Write $1=e^{i2n\pi}$. Then, we have

$$(1)^{1/5}=e^{i2n\pi/5}$$

for $n=0, \pm 1, \pm 2$. Then, the five solutions are $1$, $e^{\pm i2\pi/5}$, and $e^{\pm i4\pi/5}$.

The equation $z^4 = -256$ has four roots, $4e^{i\pi(1/4 + k/2)}$, $0\le k < 4$. The fifth roots of $1$ are $$e^{2ik\pi/5}, \qquad 0 \le k < 5.$$

• Happy Holidays. There appears to be a typo (i.e., $k/2$ instead of $k$) in the four roots of $-256$. - Mark – Mark Viola Dec 29 '15 at 19:37

The square roots of $-256$ are $16i$ and $-16i$. This is correct. Then you need to apply the square root again to these two numbers, but this will give you four solutions. The numbers $16i$ and $-16i$ have two square roots each.
As for the second problem, you need to figure out what $5$ numbers to the fifth power equal $1$. These $5$ numbers are called the fifth roots of unity: Find $z$, such that $z^5 = 1$.
It may help to write $z$ in polar form, so $z = \cos(x) + i\sin(x)$. Then $z^5$ can be found by De Moivre's formula. You get $z^5 = \cos(5x) + i\sin(5x)$, and you want this to be equal to $1$.