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.

  • 2
    $\begingroup$ Nope. Quaternions are defined in such a way that $j^2 = -1$ too. $\endgroup$ Feb 5, 2014 at 19:00
  • 8
    $\begingroup$ I'm not sure the other answers left that clear, but the square root(s) of i are the complex numbers $\frac{1 \pm i}{\sqrt{2} }$, hence you don't need to add another dimension to your field (such as j). $\endgroup$ Feb 5, 2014 at 19:02
  • $\begingroup$ Nope. Quaternions are defined in such a way that j2=−1 too right, I just meant it, it should be j2=i $\endgroup$
    – kenn
    Feb 5, 2014 at 19:07
  • $\begingroup$ And you can find the general formula for roots of complex numbers here math.stackexchange.com/questions/3315/… , for instance. $\endgroup$ Feb 5, 2014 at 19:10
  • 1
    $\begingroup$ "Deeper meanings and usage as in quaternions theory." No, $j$ and $k$ exist in quaternions because there are three "imaginary" units rather than just one as in complex numbers. They are there because they are equally part of the definition of quaternions along with the quaternion $i$, not because they were needed to solve some equation that couldn't be solved with a linear combination of $1$ and $i$. $\endgroup$
    – David K
    Dec 15, 2014 at 17:04

8 Answers 8


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.

  • $\begingroup$ Thank you for your reply. Your approach is closed to an answer to my question. What I need to know if those numbers have deeper meanings and usage. $\endgroup$
    – kenn
    Feb 5, 2014 at 19:27
  • $\begingroup$ Which do you mean by "those numbers" exactly? $\endgroup$ Feb 5, 2014 at 19:42
  • $\begingroup$ I mean i, j, k, ... according to my convention :) $\endgroup$
    – kenn
    Feb 5, 2014 at 19:48
  • 1
    $\begingroup$ As I understand it, the notation $i$ was chosen to stand for "imaginary unit" (or something like that). I don't think there is any deeper meaning, there. As far as I know, there is no standard usage of $j,k,$ etc as particular complex numbers (except that electrical engineers often use $j$ instead of $i,$ since they already use $i$ for something else). $\endgroup$ Feb 5, 2014 at 19:55

Square root of number $-1$ defined as $i$

That's the root of the problem: that people are taught that $i$ is "defined to be" something, and not that it comes naturally from the geometry and arithmetics of the problem. Because if something is "defined to be" this and that, it is automatically made non-debatable and non-explainable.

If numbers were discovered in the correct order, and in the correct way, then you would see why $i$ is what it is, because it naturally comes out of arithmetics. Here's how you can discover $i$ yourself, already with its proper geometric interpretation:

  1. A ratio of two ideas is a comparison of their properties, that is, what one needs to do with one of them to get the other. It is denoted as $a:b$.
    For example a ratio of $4:2$ tells us that you need to repeat the number $2$ twice to get the number $4$.
  2. A proportion of two (or more) ratios is when you put them into an equation, like this: $a:b = c:d$. It says that these ratios are analogous: Whatever you do with $a$ to get $b$, you do the same with $c$ to get $d$.
    Notice how multiplication is a proportion made of factors: It compares a number to its unit to figure out how that unit has to be transformed to get that number. Then it uses the other factor as a different unit, and does the same with that unit to get the product. For example, when you calculate $2\cdot3$ to get $6$, you actually do this:
    $2:1 = x:3$
    which you can put in words as: "Whatever you do with your standard unit ($1$) to get the first factor ($2$), do the same with your new unit ($3$) to get your answer ($x$)". So what do you need to do with $1$ to get $2$? You need to repeat this unit step twice. So do the same with your "bigger step", $3$: when you make two such steps of length $3$, you'll be $6$ standard unit steps farther. And that's how $2\cdot3=6$.
    As a bonus, you can now see the denominators of fractions as "units" which are repeated as many times as their numerators say.
  3. A continuous proportion, also called geometric progression, is when the denominator of the first ratio is a numerator of the second ratio: $a:b = b:c$, which can also be denoted $a:b:c$. The number $b = (a\cdot c)^{\frac{1}{2}}$ is then called geometric mean of the two extremes $a$ and $c$.
    Notice how $b$ also represents the square root of those two numbers. If you make one of them your unit, you can calculate square roots of the other number this way. But this also means that the process of taking the square root is nothing else than a continuous proportion (or geometric progression) from your standard unit ($1$), through your square root(s) ($x$), to the number $c$ you're taking the square root from: $1:x:c$.

Now we have everything we need to discover $i$ (and $-1$ along the way, too!):

What is the answer to the following geometric progression?:

$$1:x = x:1$$

Or in words: You do something with your unit ($1$) to get $x$. What do you need to to with that $x$ to get your unit ($1$) back?

The obvious answer is: Just do nothing. Let $x = 1$ (the number $1$ is the number which doesn't change any other number when multiplied with it; it is a multiplicative identity). Let's check: Substitute $1$ as $x$, and you'll get:

$$1:(1) = (1):1$$

But there's one more answer to this puzzle: You can turn your unit by $180°$, making it to operate in the opposite direction. That is, the other answer is $x = -1$, which is the opposite of $1$. Let's check:

$$1:(-1) = (-1):1$$

The proportion above says that whatever you do with your unit ($1$) to get $-1$ (which is turning it by $180°$), you do the same with the $-1$ obtained this way (that is, also rotate it by $180°$) to get your standard unit $1$ back. And this is true: Turning your unit twice by $180°$ will turn it by $360°$, which is the same as turning it by $0°$, or not turning it at all. Doing an operation twice is the same as raising to the second power, so what it says is that $1\cdot1 = 1^2 = 1$ and also $(-1)\cdot(-1) = (-1)^2 = 1$. Both $1$ and $-1$ are then square roots of unity. You can also see that if you didn't know negative numbers, you could discover them with geometric progressions: $1$ is its own geometric mean, but also $-1$ is a geometric mean between $1$ and $1$ (or half-way in between them in multiplication).

In the same way, you can discover $i$, the imaginary unit. Just ask yourself the following question: What is the geometric mean between $1$ (your standard unit) and the newly discovered $-1$? Or in symbols:

$$1 : x = x : -1$$

To turn $1$ into $-1$, you need to literally turn it by $180°$. But here you need to somehow make half of that turn, twice. And what is the half-turn of $180°$? Of course, it is $90°$! :) So your $x$, which is a square root of $-1$ (that is, an operation you need to apply twice on your unit to turn it into $-1$), turns out (huh :P) to be at the $90°$ to your standard unit. You can then call it an imaginary unit as it is commonly being called. Let's check:

$$1 : \sqrt{-1} = \sqrt{-1} : 1$$, or in modern notation: $$1 : i = i : -1$$

But remember that there are always two answers to such geometric progressions. The other one is, in this case, the opposite of $i$, that is, $-1$. Let's check:

$$1 : -\sqrt{-1} = -\sqrt{-1} : 1$$, or in modern notation: $$1 : -i = -i : -1$$

which is also true, and you can also verify it geometrically: multiplying by $-i$ is rotating by $270°$ (or, as I prefer to see it: by $-90°$, that is, by $90°$ in the opposite direction).

So you can conclude that:

$$i^2 = -1$$, or:

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

Now to your original question:

then what is the square root of complex number $i$?

To find out what are the square roots of $i$, you need to do exactly the same thing you did up to this point: You need to find a geometric mean between $1$ and $i$. Or, in symbols:

$$1 : x = x : i$$

That is, you need to do something to your unit to get that number, and then do the same again with that number to get $i$. What operation, when applied twice, rotates by $90°$? The most obvious answer is: just rotate by half of that angle, which is $45°$, or $\frac{\pi}{4}$. You don't change the length of your unit ($i$ is also a unit long), only its direction. So the answer is a number which is your unit rotated by $45°$ and still a unit long. If you were to get it by addition, you'd have to move $\frac{1}{\sqrt{2}}$ in the direction of your unit, and then another $\frac{1}{\sqrt{2}}$ in the direction perpendicular to it (that is, in the direction of your imaginary unit $i$). So the answer in rectangular/additive form is: $\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$. Or in geometry:

Square root of $i$, or fourth root of $-1$

Of course this is not the only answer! Can you find the others? ;) I'll leave it as an exercise for you. Now, when you know how it works, I'm sure you'll be able to find them.

I would say it should be $j$ as logic suggests

WHAT logic suggests that?

Sure, you can "define" $\sqrt{i} = \frac{1 + i}{\sqrt{2}}$ to be called $j$, and then of course $j\cdot j = j^2 = i$, but this has nothing to do with quaternions: your $j$ will NOT be a quaternion, just a complex number.

but it's not defined in quaternion theory in that way, am I wrong?

Yes, you are wrong. With quaternions you have:

$$i^2 = j^2 = k^2 = i \cdot j \cdot k = -1$$

so the quaternion unit $j$ is a different animal than your $j$: your $j^2 = i$, and the quaternion $j^2 = -1$.

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

Blurting out whatever definitions you like is always easy. But the real art is to make definitions which make sense and understand what meaning hides behind them.


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


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

  • $\begingroup$ Thank you for the reply, I know that proof, 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 ... $\endgroup$
    – kenn
    Feb 5, 2014 at 18:51
  • 1
    $\begingroup$ Because there is no need to add a "new" number to talk about the square root of $i$, since the complex numbers already include it. Much in the same way as asking: "Why not defining a new number $\alpha$ such that $\alpha^2 = \sqrt{2}$?" -Well, because $\sqrt[4]{2}$ already does the job and it's a plain real number. $\endgroup$ Feb 5, 2014 at 19:04

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.

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}.$

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 $\ \ \ \dfrac{i-1}{i\sqrt{2}} = \dfrac{1+i}{\sqrt{2}}$

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.

  • $\begingroup$ Thank you for your answer and formulas you shared, they are very instructive. My question is rather related to nomenclature of quaternions theory, I wondered if i, j, k .. represent different dimensions. $\endgroup$
    – kenn
    Feb 5, 2014 at 19:21
  • $\begingroup$ @kenn Glad to hear the answer was helpful. It is almost always best to ask separate questions in a new question. $\endgroup$ Feb 5, 2014 at 19:24
  • $\begingroup$ @BillDubuque I tried your nice formula for $\sqrt[3]{-4+2\sqrt{-1 } }$ but it seems doesnt work.can you explain why? $\endgroup$ Mar 5, 2021 at 10:17
  • $\begingroup$ @program_craft The formula denests square roots, but that's a cube root. $\endgroup$ Mar 5, 2021 at 15:02
  • $\begingroup$ @BillDubuque Have any idea about cube root? I stucked link $\endgroup$ Mar 5, 2021 at 15:07

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



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)


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?

  • $\begingroup$ Thank you for the answer and correction, I edited the title of my question. $\endgroup$
    – kenn
    Feb 5, 2014 at 21:31

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.

  • $\begingroup$ Please use MathJac to render the math thank you. $\endgroup$ Mar 5, 2021 at 17:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .