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Square root of i?

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

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

\(1, -1, i, -i, \sqrt{i}, -\sqrt{i} \sqrt{-i} \text{and} -\sqrt{-i}.\)

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

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

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

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

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

Note by Eamon Gupta
1 year, 2 months ago

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Finding the square roots of \(i\) is the same thing as solving the equation \(x^2=i\) in complex numbers.

Now, what you did is raise both sides of this equation to the fourth power to get \(x^8=1\). While it is true that if \(x^2=i\), then \(x^8=1\), 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 \(x=4\), then \(x^2=16\). But \(x^2=16\) implies \(x\) is equal to either \(4\) or \(-4\). So squaring gave you something (\(x=-4\)) that's not a solution to the orginal equation.

The fundamental theorem of algebra ensures that an \(n\)-degree polynomial with complex coefficients can have at most \(n\) 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 · 1 year, 2 months ago

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@Mursalin Habib As \(x=4\) and \(x^2=16\) aren't same, \(x^2=i\) and \(\sqrt{i} \) aren't same too.

Solving \(x^2=i\), we get \(x=e^{\frac{i\pi} {4}}=\sqrt[4]{-1},e^{-\frac{3i\pi}{4}}\).

But \(\sqrt{i} \) means principal root of \(i\) which is \(\sqrt[4]{-1}\) Md Omur Faruque · 1 year, 2 months ago

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@Md Omur Faruque 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 \(\sqrt{z}=\sqrt{|z|}\frac{z+|z|}{|z+|z||}\)) 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 \(e^{i\frac{\pi}{2}}\) [even though they didn't state it explicitly] and I was aiming to address that. Mursalin Habib · 1 year, 2 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 · 1 year, 2 months ago

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@Mursalin Habib Ok thanks for clearing that up. I'll look into some quaternions... Eamon Gupta · 1 year, 2 months ago

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It can be use in hard forms Shrey Kannad · 1 year, 2 months ago

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