5

Where $i$ is the square root of negative one. And is there a generalization of the $n$th root of $i$? Also how would this look graphically on the real number axis? Thanks

choco_addicted
  • 8,786
  • 8
  • 24
  • 53
Amour
  • 395
  • Have you heard of de Moivre's theorem, or Euler's formula? – beep-boop Oct 08 '14 at 23:20
  • 3
    There's no such thing as the square roots of non-real numbers. Read this answer to see how to get all the $n^{\text{th}}$ roots of any complex number, for all natural number $n$. – Git Gud Oct 08 '14 at 23:21
  • Which square root of $i$? There are two, both of which are non-real, and (unlike nonnegative real numbers) there is no obvious choice for "the" square root most complex numbers. – Cameron Buie Oct 08 '14 at 23:22
  • I've answered this before here http://math.stackexchange.com/questions/1348934/taking-the-square-root-of-an-imaginary-number, I hope you check out my answer as I think it's what your looking for. It's the fourth answer. – Ahmed S. Attaalla Feb 06 '16 at 02:00

4 Answers4

5

If $\;w=a+bi\in\Bbb C\;$, then $\;z=x+yi\;$ fulfills $\;z^2=w\;$ iff

$$z^2=x^2-y^2+2xyi =a+bi\iff \begin{cases}x^2-y^2=a\\{}\\2xy=b\end{cases}$$

The above easy system has at least one solution and at most two different ones, and you'll have to be careful with the signs (pay attention in particular to the sign of $\;b\;$ .

Now, for your question only take $\;a=0\,,\,\,b=1\;$

Timbuc
  • 34,191
4

To answer your question directly, one choice of $n$th root is as follows:

A square root of $i$ is given by $$ z = \frac 1{\sqrt {2}} (1+i) $$ Graphically, this is a $45^\circ$ ray of length $1$ from the origin of the complex plane.

An $n$th root of $i$ is given by $$ z = \cos(\pi/(2n)) + i\sin(\pi/(2n)) $$ Graphically, this is a $(90/n)^\circ$ ray of length $1$ from the origin of the complex plane.

Ben Grossmann
  • 225,327
  • Beautiful, I really like how "i" has a real and imaginary part to it and that you can see it by simply graphing it! – Amour Oct 08 '14 at 23:40
3

If you learned the trigonometric form of complex numbers, the problem is very easy to solve that way.

Just solve $$[R(\cos(\theta)+i\cos(\theta))]^2=i$$

Otherwise, let $z=a+bi$. Then $$z^2=i \Leftrightarrow (a+bi)^2=i \Leftrightarrow a^2-b^2+2abi=i $$

This reduces to $$a^2=b^2$$ $$2ab=1$$

Which is easy to solve.

N. S.
  • 132,525
  • I haven't learned the trigonometric form. In your second way I'm assuming you let a^2 - b^2 equal the real part of i, which is clever. Thanks! – Amour Oct 08 '14 at 23:34
  • So is that a = b = sqrt(0.5) (positive or negative) ? –  Jan 07 '15 at 08:36
  • @paxdiablo $2ab=1$ implies that they both have the same sign. So yes, but both positive or both negative. – N. S. Jan 07 '15 at 14:11
1

According to de Moivre's formula, if we have complex number $z = r(cos\phi + isin\phi)$, then the equation $x = \sqrt[n]{z}$ has exactly $n$ different roots over a complex plane. $$\sqrt[n]{z} = \sqrt[n]{r}(\cos \frac{\phi + 2\pi k}{n} + i\sin \frac{\phi + 2\pi k}{n}), k = 0, 1, ... n-1$$ So, for example $$(\sqrt{i})_1 = \cos(\pi + \pi/4) + i\sin(\pi + \pi/4) = -\sqrt{2}/2-i\sqrt{2}/2$$ $$(\sqrt{i})_2 = \cos(2\pi + \pi/4) + i\sin(2\pi + \pi/4) = \sqrt{2}/2+i\sqrt{2}/2$$

These values obviosly can't be shown on real number axis but can be presented as dots on a complex plane: if $z = a + bi$, then you should take $a$ for $x$ coordinate and $b$ for $y$.
There's also an insteresting fact concerning graphical representation: roots of a complex number are always uniformly placed on a circle of radius $\sqrt[n]{r}$ and center in $(0,0)$ with $2\pi / n$ arc between every two adjacent roots.

Andrei Rykhalski
  • 1,325
  • 9
  • 14