What is the square root of complex number i?

Question

Square root of number -1 defined as i, then what is the square root of complex number i?, I would say it should be j as logic suggests but it's not defined in quaternion theory in that way, am I wrong?

EDIT: my question is rather related to nomenclature of definition, while square root of -1 defined as i, why not j defined as square root of i and k square root of j and if those numbers have deeper meanings and usage as in quaternions theory.

Answer 1

Unfortunately, this cannot be answered definitively. In fact, every non-zero complex number has two distinct square roots, because $-1\ne1,$ but $(-1)^2=1^2.$ When we are discussing real numbers with real square roots, we tend to choose the nonnegative value as "the" default square root, but there is no natural and convenient way to do this when we get outside the real numbers.

In particular, if $j^2=i,$ then putting $j=a+bi$ where $a,b\in\Bbb R,$ we have $$i=j^2=(a+bi)^2=a^2-b^2+2abi,$$ so we need $0=a^2-b^2=(a+b)(a-b)$ and $2ab=1.$ Since $0=(a+b)(a-b),$ then $a=\pm b.$ If we had $a=-b,$ then we would have $1=2ab=-2b^2,$ but this is impossible, since $b$ is real. Hence, we have $a=b,$ so $1=2ab=2b^2,$ whence we have $b=\pm\frac1{\sqrt{2}},$ and so the square roots of $i$ are $\pm\left(\frac1{\sqrt{2}}+\frac1{\sqrt{2}}i\right).$

I discuss in my answer here that $i$ is defined as one of two possible numbers in the complex plane whose square is $-1$ (it doesn't actually matter which, as far as the overall structure of the complex numbers is concerned). Once we've chosen our $i,$ though, we have fixed which "version" of the complex numbers we're talking about. We could then pick a canonical square root of $i$ (and call it $j$), but there's really no point. Once we've picked our number $i,$ we have an algebraically closed field, meaning (incredibly loosely) that we have all the numbers we could want already there, so we can't (or at least don't need to) add more, and there's no particular need to give any others of them special names.

Answer 2

Since $$i=e^{i\pi/2}$$ then $$e^{i(\pi/4+k\pi)},\;k=0,1$$ are the two square roots of $i$.

Answer 3

I think this would be easier to see by writing $i$ in its polar form,

$$i=e^{i\pi/2}$$

This shows us that one square root of $i$ is given by $$i^{1/2}=e^{i\pi/4} $$

Answer 4

$$\Big(\frac{i+1}{\sqrt{2}}\Big)^2=\frac{(i+1)^2}{2}=\frac{1-1+2i}{2}=i$$

In general $\mathbb{C}$ is algebraically closed (fundamental theorem of algebra). So square root of every complex number is a complex number. You can find square root of $a+ib$ by solving $$(x+iy)^2=a+ib$$ (Eqaute the real and imaginary parts)

Answer 5

Denesting $\sqrt{a+b\sqrt{n}},\ $ here $\,\sqrt{\sqrt{-1}}=\sqrt i,\, $ can be done by a simple formula explained here.

Simple Denesting Rule $\rm\ \ \, \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{#0a0}{divide\ out}\ \sqrt{trace} $

Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $

and, $ $ furthermore, $\rm\ \ w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2a$

$\!\rm\color{#c00}{Any\ sign}$ is ok in norm & trace sqrts, e.g. we do both $\,\color{#c00}{\bf \sqrt{1} = \pm1}\,$ below.

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Here $\,i\:$ has norm $= 1.\:$ $\rm\ \color{blue}{Subtracting\ out}\ \sqrt{norm}\ =\, \color{#c00}{\bf -1}\ $ yields $ \ i+1,\ $ i.e. $\ i-(\bf\color{#c00}{-1})$

which has $\rm\ \sqrt{trace}\: =\: \sqrt{2}.\ \ \color{#0a0}{Dividing\ it\ out}\ $ of the above yields $ \ (i+1)/\sqrt{2}.$

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Or: $\ i\:$ has norm $\,=\, 1.\:$ $\rm\ \color{blue}{Subtracting\ out}\ \sqrt{norm}\ \,=\, \color{#c00}{\bf +1}\ $ yields $ \ i-1\:$

which has ${\rm\ \sqrt{trace}}\: =\: \sqrt{-2}\: =\: i\sqrt{2}. \rm\ \ \ \color{#0a0}{Dividing\ it\ out}\ $ yields $\ \ \ \dfrac{i-1}{i\sqrt{2}} = \dfrac{1+i}{\sqrt{2}}$

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Generally, as above, we are free to choose the $\rm\color{#c00}{sign}$ of the norm & trace square-roots as we please, e.g. so that the arithmetic is simpler, as in the first case above. This follows from the proof in the linked answer. For many further worked examples see my prior posts on denesting.

Answer 6

The answer is that

the square root of $i$ isn't defined.

That is, when we talk about the square root of something we are talking about the function $$ x \mapsto \sqrt{x}. $$ Being a function means that for each input, you have exactly one output. For real numbers the square root of a non-negative number $x$ is defined as the unique non-negative number $y$ satisfying that $y^2 = x$.

The thing is that we don't have a universally agreed upon definition of the square root of a complex number.

Now, you can of course consider the equation $y^2 = i$. This has two solutions: $$ y = e^{\frac{2\pi i}{4}} \quad \text{or} \quad y = e^{5\frac{2\pi i}{4}}. $$ If you wanted to define a square root, you would be faced with the question of which root would you pick.

So in general it isn't good to be talking about the square root as having two values because we would like to think about the square root as a function. One might talk about a square root.

For more on this, talk a look at the answers to this question: How do I get the square root of a complex number?

Answer 7

To get rid of real numbers, x^2=-y^2, because if you add a number to it's negative the result is 0, and what's left is an imaginary number. If x^2=-y^2, then x=iy. So lets plug in x=1, and y=i. (1+i)^2=1^2+2i+i^2=1+2i-1=2i. Not quite, but almost there. (sqrt1/2+sqrt1/2i)^2=(sqrt1/2)^2+2sqrt1/2sqrt1/2i+(sqrt1/2i)^2=1/2+21/2i-1/2=0+21/2i=i Therefore, the answer is sqrt1/2+sqrt1/2i or -(sqrt1/2+sqrt1/2i) or sqrt1/2-sqrt1/2i ((-i)^2=-1) or sqrt1/2i-sqrt1/2.