i\sqrt{i}

Why is i=22+22i\sqrt{i} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i

Note by Saad Haider
6 years ago

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8 votes

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i=eiπ/2i=eiπ/4=cos(π/4)+isin(π/4)=22+22i\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 - 6 years ago

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remember taking root will give both positive and negative solutions

Manish Kansal - 6 years ago

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

Joel Jablonski - 6 years ago

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Sorry about that. Fixed.

Thanks! :)

Pranav Arora - 6 years ago

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Actually, ii has two square roots, just as any real number does. These are the two roots of z2=eπi/2z^2=e^{\pi i/2}. One of the roots is eπi/4=22+22ie^{\pi i/4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i as noted previously. The other is e5πi/4=2222ie^{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 eπi/2e^{\pi i/2}.

Aaron Doman - 6 years ago

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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?

Jonathan Wong - 6 years ago

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The "principal root" of x describes the positive root, so if you're given 4 \sqrt{4} then it is asking for the positive answer which is 22 and not 2-2. 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 - 6 years ago

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@Michael Tong and for higher roots? (e.g. 25\sqrt[5]{2})

Jonathan Wong - 6 years ago

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You're missing a minus sign:

e5πi/4=2222ie^{5 \pi i / 4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i

Ton de Moree - 6 years ago

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let i\sqrt{i} = xx

we have x2x^{2} = ii

converting i to polar form, we have

x2x^{2} = cosπ2+isinπ2\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}

using Demoivre's Theorem

we have x1=cosπ4+isinπ4=22+22ix_{1} = \cos \frac{\pi}{4} + i\sin \frac{\pi}{4} = \frac{\sqrt2}{2} + \frac{\sqrt2}{2} i and x2=cos5π4+isin5π4=2222ix_{2} = \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} = -\frac{\sqrt2}{2} -\frac{\sqrt2}{2} i

Ed Mañalac - 6 years ago

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

Squaring gives (a2b2)+2abi(a^2-b^2)+2abi. It follows that 2ab=12ab=1 and a2=b2a^2=b^2.

From the second we have that a=ba=b or a=ba=-b.

Substituting in the first, we get a2=12a^2=\frac{1}{2} or a2=12a^2=-\frac{1}{2}

So a=±122a=\pm \frac{1}{2} \sqrt{2} or a=±122ia=\pm \frac{1}{2} \sqrt{2}*i.

and b=±122b=\pm \frac{1}{2} \sqrt{2} or b=122ib=\mp \frac{1}{2} \sqrt{2}*i.

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

122+122i \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2}*i and 122122i- \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{2}*i .

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

Ton de Moree - 6 years ago

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