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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. · 1 year, 2 months ago

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 $$\sqrt[4]{-1}$$ · 1 year, 2 months ago

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. · 1 year, 2 months ago

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