Square root of i?

What is wrong with this proof or is it in fact it is valid?

i=(i)12=(i)48=i48=18= \large \sqrt{i} = (i)^{\frac{1}{2}} = (i)^{\frac{4}{8}} = \sqrt[8]{i^{4}} = \sqrt[8]{1} =

1,1,i,i,i,iiandi.1, -1, i, -i, \sqrt{i}, -\sqrt{i} \sqrt{-i} \text{and} -\sqrt{-i}.

Does that mean that all 8 solutions are solutions of i\sqrt{i}?

I know that i\sqrt{i} can be expressed as ±12×(1+i)\pm\frac{1}{\sqrt{2}}\times(1+i) but my questions are:

1) Which of the 8 solutions are invalid and why are they invalid?

2) Is it possible of i\sqrt{i} to have more than 2 solutions? If so, what are they?

3) Is it possible for i\sqrt{i} to have any solutions where there is a real part (in addition to the 2 given)?

Note by Eamon Gupta
5 years, 12 months ago

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Finding the square roots of ii is the same thing as solving the equation x2=ix^2=i in complex numbers.

Now, what you did is raise both sides of this equation to the fourth power to get x8=1x^8=1. While it is true that if x2=ix^2=i, then x8=1x^8=1, but the converse is not always true. In other words, when you are raising the equation to a power, you're introducing extraneous solutions.

Take this for example. If x=4x=4, then x2=16x^2=16. But x2=16x^2=16 implies xx is equal to either 44 or 4-4. So squaring gave you something (x=4x=-4) that's not a solution to the orginal equation.

The fundamental theorem of algebra ensures that an nn-degree polynomial with complex coefficients can have at most nn complex roots. So that means a complex number can not have more than two complex square roots. But higher dimensional numbers (like quaternions) are a different story.

Mursalin Habib - 5 years, 11 months ago

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Ok thanks for clearing that up. I'll look into some quaternions...

Eamon Gupta - 5 years, 11 months ago

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As x=4x=4 and x2=16x^2=16 aren't same, x2=ix^2=i and i\sqrt{i} aren't same too.

Solving x2=ix^2=i, we get x=eiπ4=14,e3iπ4x=e^{\frac{i\pi} {4}}=\sqrt[4]{-1},e^{-\frac{3i\pi}{4}}.

But i\sqrt{i} means principal root of ii which is 14\sqrt[4]{-1}

MD Omur Faruque - 5 years, 11 months ago

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It is possible to extend the definition of a principle square root to an arbitrary complex number, but it's usually not done because it is not as useful as the principal square root of a non-negative real. (The definition you're using is something like this: the principal square root is the one with the positive real part or the one that lies in the right-complex-half-plane, or z=zz+zz+z\sqrt{z}=\sqrt{|z|}\frac{z+|z|}{|z+|z||}) That's why the principal square root operator, \sqrt{\quad} is generally not used on a general complex number. It doesn't offer us much to talk about a particular square root of a general complex number.

The OP was actually looking for the square roots (plural) of eiπ2e^{i\frac{\pi}{2}} [even though they didn't state it explicitly] and I was aiming to address that.

Mursalin Habib - 5 years, 11 months ago

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@Mursalin Habib I am really interested to contact you. If you don't mind can I have your email address or something?

MD Omur Faruque - 5 years, 11 months ago

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It can be use in hard forms

Shrey Kannad - 5 years, 11 months ago

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