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The square root of i is $\frac{\sqrt{2} + i \sqrt{2}}{2}$. But how is it valid to use a number in expressing the square root of that number?

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    $\begingroup$ The square root of $i$ is not defined as $\frac{\sqrt2+i\sqrt2}{2}$: it just is that number. $\endgroup$ – Mariano Suárez-Álvarez May 10 '13 at 3:44
  • $\begingroup$ I have fixed this. $\endgroup$ – Lee Sleek May 10 '13 at 3:56
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    $\begingroup$ A square root of 1 is 1. Surely you don't find that objectionable? $\endgroup$ – 200_success May 10 '13 at 7:01
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The set of complex numbers form an algebraically closed field. Hence, there exists solutions to $z^2=i$. You may want to look up the fundamental theorem of algebra. This guarantees that there are no rabbit holes, i.e., you do not need to move away from complex numbers. The problem of solving for square-roots, cube-roots and in general any polynomial equation is solved once and for all by the complex numbers.

In your case, you want to find a $z$ such that $z^2 = i$. If $z=x+iy$, where $x,y \in \mathbb{R}$, we get that $$z^2 = (x^2-y^2) + i(2xy) = i \implies x = \pm y \text{ and } xy = \dfrac12 \implies x=y=\dfrac1{\sqrt{2}} \text{ or } x=y= - \dfrac1{\sqrt2}$$ Hence, we get that $$i^{1/2} = \pm \left(\dfrac{\sqrt2 + i \sqrt2}{2}\right)$$ (Note that one of your solution is incorrect.)

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Square it and see. It works fine. In the complex field, you can take the square root of any number. If we use the polar representation $z=re^{i\theta}$ ($r, \theta$ real), the square roots will be $\sqrt r e^{\frac {i \theta}{2}}$ and $\sqrt r e^{\frac {i \theta}{2}+i\pi}$ This extends to $n^{\text {th}}$ roots. The $n^{\text {th}}$ root of $z=re^{i\theta}$ is $r^{\frac 1n}e^{i(\frac \theta n+\frac {2k\pi}n)}$ for $k=0,1,2,\ldots n-1$

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  • $\begingroup$ I know it works fine; I just don't think it's right that expressing it requires the very number of which we wish to take the square root. $\endgroup$ – Lee Sleek May 10 '13 at 3:36
  • $\begingroup$ @LeeSleek: First we define the complex numbers, by saying there is something called $i$ such that $i^2=-1$ and they are all the things of the form $a+bi$, with $a,b$ real. Then we prove they form a field. Now it is a fair question whether there is something in this set that can be squared to get $i$ and we answer it affirmatively. It is not circular. We are working in an environment where $i$ is defined already. $\endgroup$ – Ross Millikan May 10 '13 at 3:43
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    $\begingroup$ @LeeSleek, note that $\sqrt{1}=\pm 1$; expressing this requires the very number whose square root we are taking. $\endgroup$ – vadim123 May 10 '13 at 3:46
  • $\begingroup$ @LeeSleek: Sorry, but this sounds like an attempt to rescue the question. We define the naturals as successors of $0$ and add addition and multiplication. Then we say $2^2=4$, which uses something defined before $4$. $\endgroup$ – Ross Millikan May 10 '13 at 4:05

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