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So, we all know that $\sqrt{-1}=i$, that $\sqrt{-2}=\sqrt2i$, and so on. And that, for example, $\sqrt[3]{-27}=-3$. But I was wondering; what would $\sqrt[4]{-1}$ be? Since all $n^{th}$ roots of $-1$ that are odd result in $-1$ (Because $-1^{1,3,5,7,9,\dots}=-1$), I wondered what variation of $i$ would even roots of $-1$ give? (Like $\sqrt{-1},\sqrt[4]{-1},\sqrt[6]{-1},\dots$)

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    $\begingroup$ You're wrong: $\sqrt{-2} \neq 2 i$. $\endgroup$ – David G. Stork May 3 '18 at 1:36
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    $\begingroup$ $\sqrt{-4}=2i$. Meanwhile $\sqrt{-2}=(\sqrt{2})i$ $\endgroup$ – JMoravitz May 3 '18 at 1:45
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    $\begingroup$ Shouldn't the title be $\sqrt [4] {-1}$ as opposed to $\sqrt [4] {i}$ by the way? $\endgroup$ – Mr Pie May 3 '18 at 1:53
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    $\begingroup$ Related to my earlier points: what is the principal cubic root of 8 $\endgroup$ – JMoravitz May 3 '18 at 2:06
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    $\begingroup$ You shouldn't write $-1^3$ if you mean $(-1)^3.$ Those are two different things. $\endgroup$ – Michael Hardy May 3 '18 at 2:30
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Why would $\sqrt{-1}=i$ and not $\sqrt{-1}=-i$? There is no real reason to choose.

What we do is to enumerate the roots; there are two square roots, three cubics roots, four fourth roots, etc.

From $-1=\cos\pi+i\sin\pi$, we can deduce via De Moivre's formula (or using $-1=e^{i\pi}$), that the four fourth roots of $-1$ are $$ \cos\frac{(2k+1)\pi}4+i\sin\frac{(2k+1)\pi}4,\ \ k=0,1,2,3. $$ Explicitly, you have $$ \cos\frac\pi4+i\sin\frac\pi4=\frac1{\sqrt2}+i\,\frac1{\sqrt2}, $$ $$ \cos\frac{3\pi}4+i\sin\frac{3\pi}4=-\frac1{\sqrt2}+i\,\frac1{\sqrt2}, $$ $$ \cos\frac{5\pi}4+i\sin\frac{5\pi}4=-\frac1{\sqrt2}-i\,\frac1{\sqrt2}, $$ $$ \cos\frac{7\pi}4+i\sin\frac{7\pi}4=\frac1{\sqrt2}-i\,\frac1{\sqrt2}. $$ Note that similarly there is no single cubic root of $-1$, but three of them: $$ \cos\frac{(2k+1)\pi}3+i\sin\frac{(2k+1)\pi}3,\ \ k=0,1,2. $$

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  • $\begingroup$ Stupid question: Why are there four fourth roots? Or three cubic roots? There’s only one answer to, for example, the cubic root of -27, or the fifth root of -32. (I’m in Grade 7, so some things are too advanced for me.) $\endgroup$ – Detmondyou May 3 '18 at 1:43
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    $\begingroup$ Not stupid at all. Any complex number $c$ has four fourth roots. One way to see it is that $1$ has four roots (that can be obtained as in my answer, they are equally spaced on the unit circle), and they are of the form $1,r,r^2,r^2$. So if now $x^4=c$, then $(rx)^4=r^4x^4=1c=c$. $\endgroup$ – Martin Argerami May 3 '18 at 1:48
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    $\begingroup$ @Detmondyou See the Fundamental theorem of Algebra. $\endgroup$ – Simply Beautiful Art May 3 '18 at 1:52
  • $\begingroup$ @SimplyBeautifulArt: my first version of the comment was about it. But the FTA will not tell you that the roots are not repeated. It will only tell you that there is a root. $\endgroup$ – Martin Argerami May 3 '18 at 1:57
  • $\begingroup$ Indeed, but it may help to point out that an n degree polynomial has n (possibly repeating) roots. It seems the OP is not very aware of that fact. $\endgroup$ – Simply Beautiful Art May 3 '18 at 2:01
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We will use Euler's Identity, $$e^{i\pi} + 1 = 0,\tag1$$ or in particular, $$\begin{align} e^{i\theta} &= \cos\theta + i\sin\theta \\ &= \text{cis}\,\theta.\end{align}\tag2$$


From $(1)$, we get that $e^{i\pi} = -1$, thus $$\begin{align} \left(e^{(i\pi)/2}\right)^{1/2} &= e^{(i\pi)/4} \\ &= \sqrt [4] {-1}.\end{align}$$ Now, using $(2)$, by substituing $\theta = \dfrac{\pi}{4}$, we obtain the following result: $$\begin{align} \sqrt [4] {-1} &= \cos \left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) \\ \\ &= \frac{1}{\sqrt{2}} + i\left(\frac{1}{\sqrt{2}}\right) \\ \\ &= \boxed{ \ \frac{1}{\sqrt{2}}\left(1+i\right). \ }\end{align}$$ See what pattern you can find now with the sequence, $\sqrt{-1}, \sqrt [4] {-1}, \sqrt [6] {-1},\ldots$

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Method 1 Hint: Consider $$z^4=i$$ where $$z=a+bi$$ with $a,b\in\mathbb{R}$.

Expand and compare coefficients. (Tedious)


Method 2:

Consider $$i=i \cdot1^k= \exp\left({\frac\pi2i}\right)\cdot\left(\exp(2\pi i )\right)^k=\exp\left({\frac\pi2i+2k\pi i}\right)$$ for $k\in\mathbb{Z}$

Let $$z=\exp({\theta i})$$ such that $$z^4=i$$ Then use the relation $$\exp(\theta i)=\cos\theta+i\sin\theta$$

So \begin{align} \exp(4\theta i)&= \exp\left({\frac\pi2i+2k\pi i}\right)\\ 4\theta i&= \frac\pi2i+2k\pi i\\ \theta&=\frac\pi8+\frac\pi2k\\ \end{align}

Hence, the fourth roots of $i$ are $$z=\cos\theta+i\sin\theta$$with $$\theta= -\frac{7\pi} 8,-\frac{3\pi}8,\frac\pi8,\frac{5\pi}8$$ for $\theta\in(-\pi,\pi]$

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    $\begingroup$ This answer is also good, just not as clear as Martin Argerami. $\endgroup$ – Detmondyou May 3 '18 at 2:13
  • $\begingroup$ Just a mention one needs to find $z^4=-1$ not $z^4=i$, so for the first one you need to solve $z^2=i$ and $z^2=-i$ which isn't that tedious $\endgroup$ – kingW3 Jun 19 '18 at 21:57

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