Actually, \(i\) has two square roots, just as any real number does. These are the two roots of \(z^2=e^{\pi i/2}\). One of the roots is \(e^{\pi i/4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) as noted previously. The other is \(e^{5\pi i/4}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\). Doubling the argument of both of these complex numbers and squaring the modulus indeed gives \(e^{\pi i/2}\).

The "principal root" of x describes the positive root, so if you're given \( \sqrt{4}\) then it is asking for the positive answer which is \(2\) and not \(-2\). but when dealing with complex numbers it gets.. well, more complex. Neither root is more "important" than the other, so you don't tend to worry about it as much.

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TopNewest\(\displaystyle i=e^{i\pi/2} \Rightarrow \sqrt{i}=e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\)

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remember taking root will give both positive and negative solutions

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I completely agree though you forgot to put in the \(i\) until the end: \(i=e^{\pi i/2}\) Voted up. :D

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Sorry about that. Fixed.

Thanks! :)

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Actually, \(i\) has two square roots, just as any real number does. These are the two roots of \(z^2=e^{\pi i/2}\). One of the roots is \(e^{\pi i/4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) as noted previously. The other is \(e^{5\pi i/4}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\). Doubling the argument of both of these complex numbers and squaring the modulus indeed gives \(e^{\pi i/2}\).

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You're missing a minus sign:

\(e^{5 \pi i / 4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\)

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There is some method with which we agree on which root we're talking about with radicals, making radicals ambiguous but fractional exponents not so.

Sadly, I don't know exactly how we do decide. Could anyone enlighten me?

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The "principal root" of x describes the positive root, so if you're given \( \sqrt{4}\) then it is asking for the positive answer which is \(2\) and not \(-2\). but when dealing with complex numbers it gets.. well, more complex. Neither root is more "important" than the other, so you don't tend to worry about it as much.

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let \(\sqrt{i}\) = \(x\)

we have \(x^{2}\) = \(i\)

converting i to polar form, we have

\(x^{2}\) = \(\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}\)

using Demoivre's Theorem

we have \(x_{1} = \cos \frac{\pi}{4} + i\sin \frac{\pi}{4} = \frac{\sqrt2}{2} + \frac{\sqrt2}{2} i\) and \(x_{2} = \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} = -\frac{\sqrt2}{2} -\frac{\sqrt2}{2} i\)

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\(\sqrt{i}\) is a complex number, so it can be written in the form \(a+bi\) where \(a\) and \(b\) are real numbers.

Squaring gives \((a^2-b^2)+2abi\). It follows that \(2ab=1\) and \(a^2=b^2\).

From the second we have that \(a=b\) or \(a=-b\).

Substituting in the first, we get \(a^2=\frac{1}{2}\) or \(a^2=-\frac{1}{2}\)

So \(a=\pm \frac{1}{2} \sqrt{2}\) or \(a=\pm \frac{1}{2} \sqrt{2}*i\).

and \(b=\pm \frac{1}{2} \sqrt{2}\) or \(b=\mp \frac{1}{2} \sqrt{2}*i\).

Using these in our original expression \(a+bi\), we get only two distinct solutions:

\( \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2}*i \) and \(- \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{2}*i \).

I think usually the first is chosen as THE root of \(i\), but only because it's closest to \((1,0)\).

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