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\(\sqrt{i}\)

Why is \(\sqrt{i} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\)

Note by Saad Haider
3 years, 10 months ago

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\(\displaystyle i=e^{i\pi/2} \Rightarrow \sqrt{i}=e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) Pranav Arora · 3 years, 10 months ago

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@Pranav Arora remember taking root will give both positive and negative solutions Manish Kansal · 3 years, 10 months ago

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@Pranav Arora I completely agree though you forgot to put in the \(i\) until the end: \(i=e^{\pi i/2}\) Voted up. :D Joel Jablonski · 3 years, 10 months ago

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@Joel Jablonski Sorry about that. Fixed.

Thanks! :) Pranav Arora · 3 years, 10 months ago

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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 \(e^{\pi i/2}\). Aaron Doman · 3 years, 10 months ago

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@Aaron Doman You're missing a minus sign:

\(e^{5 \pi i / 4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\) Ton De Moree · 3 years, 10 months ago

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@Aaron Doman 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? Jonathan Wong · 3 years, 10 months ago

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@Jonathan Wong 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. Michael Tong · 3 years, 10 months ago

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@Michael Tong and for higher roots? (e.g. \(\sqrt[5]{2}\)) Jonathan Wong · 3 years, 10 months ago

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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 \(x_{2} = \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} = -\frac{\sqrt2}{2} -\frac{\sqrt2}{2} i\) Ed Mañalac · 3 years, 10 months ago

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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 \((1,0)\). Ton De Moree · 3 years, 10 months ago

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