I am reading complex analysis by Gamelin and I am having trouble understanding the square root function.

The principal branch of $\sqrt{z}$ ( $f_1(z)$ ) is defined as $|z|^{\frac 1 2} e^{\frac{i \operatorname{Arg}(z)}{2}}$ for $z \in \mathbb{C} - (-\infty,0]$ and $f_2(z)$ is defined as $-f_1(z)$ where $\operatorname{Arg}(z) \in (-\pi , \pi]$

By this definition, what is $\sqrt{-r} $ where $r$ is a non negative real number? Of course the answer is $i\sqrt{r}$ but the definition of square root functions doesn't apply here

What is $i$ then? $i:=\sqrt{-1}$ but how? We didn't define the square root function for negatives but we still use $i$ to define complex numbers. Shouldn't the definition of square root function taught before defining $i$?

and defining $i^2=-1$ without define $i$ as either $\pm \sqrt{-1}$ is also very strange because we want extended function to be continuous and choosing $\pm \sqrt{-1}$ will make this impossible although $i$ must be one of the two (after defining the square root of negatives).

Why do we even care so much about continuity? (I am sure there is an answer is the rest of the book but I am still a beginner), all of this confusion will disappear if we just made $i:= \sqrt{-1}$, we seem all to agree to use $i:= \sqrt{-1}$ in solving problems involving complex numbers,even wolfram alpha uses that and every high school math book and some engineering math books (I know the last know types of books lacks rigorousness in general but what I am saying is is seems normal to just define $i$ that way ), so why all of that confusion?

I asked some of my graduate friends and all of them told me the they just define $i:=\sqrt{-1}$ without any of these complications.

Now I am more confused, do we define $i:=\sqrt{-1}$ or not? If we do what did I get wrong from the definition of principal root?

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    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Xander Henderson
    Commented Mar 4 at 18:32
  • $\begingroup$ Hadn't realized it before, but seems there is a bit of circular reasoning. $i$ is used to define the main branch, and other branches of the square root function, but $i$ is the square root of $-1$. I think the historical development of complex arithmetic might help. Cardano discovered them in the context of solving the cubic. Apparently the cubic formula will require taking the square root of negative numbers even if all roots are real. This helps tease out conjugacy and other defining properties. $\endgroup$ Commented Mar 14 at 19:59
  • $\begingroup$ @TurlocTheRed: There is no circular reasoning if one is just a little bit careful. $\endgroup$ Commented Mar 14 at 22:14

6 Answers 6


An alternative way to construct the complex plane $\mathbb{C}$ is to take the vector space $\mathbb{R}^2$ and equip it with a multiplication function $\ast: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2$ defined by:

$$(x_1, y_1) \ast (x_2, y_2) = (x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1)$$

You can check that is satisfies the usual 'nice' properties of multiplication (associativity, commutativity, etc.), so the name is justified.

We can now denote a vector in this space by $x + yi := (x, y)$, where $i$ here really just means the "second coordinate direction." From here, the trick is to think of $x + yi$ as a "number" (which we'll call a complex number), and see how the normal multiplication of numbers aligns with our multiplication operation. You'll find that

$$(x_1 + iy_1)(x_2 + iy_2) = (x_1x_2 + i^2 y_1 y_2) + i(x_1 y_2 + x_2 y_1)$$

So we have compatibility if we interpret the symbol $i^2$ as $-1$. It's notational convenience and nothing more, it does not say that $i = \sqrt{-1}$, just that thinking that way is symbolically consistent with the above construction.

I find it similar to treating differentials as algebraic objects when moving from integral calculus to ODEs. It's a highly convenient abuse of notation.


$\DeclareMathOperator{\Arg}{Arg}$The times I taught complex analysis using the (good, free) book of Beck, Marchesi, Pixton, and Sabalka, there was a nitpicky but useful handling polar angle:

  • The principal branch of argument $\Arg$ is defined for all non-zero complex numbers, and $-\pi < \Arg \leq \pi$. One must take care that $\Arg$ is discontinuous on the negative real axis, but on the negative real axis $\Arg = \pi$.

  • The non-positive real axis is explicitly excluded from domains of differentiable functions whose definition involves $\Arg$: $\log$, roots, other non-integer power functions.

In the question, the negative real axis is explicitly excluded from the domain of the principal square root (second bullet point); consequently, as José says, with this definition $\sqrt{-1}$ (the principal square root of $-1$) is undefined.

If we are unconcerned about the discontinuity of the principal square root on the negative real axis, however (first bullet point), then for all positive real $r$ we have $\sqrt{-r} = re^{i\pi/2} = ir$.

As for developing the complex square root before defining $i$, that's not necessary, and arguably not feasible. (What are we taking square roots of before $i$ is defined?)

Nowadays mathematicians explicitly or implicitly define a complex number to be an ordered pair $(a, b)$ of real numbers, equipped with operations of addition and multiplication consistent with the formal identification $(a, b) \leftrightarrow a + bi$ subject to $i^{2} = -1$. In this language, and subject to the first convention above, $$ \sqrt{-1} = \sqrt{(-1, 0)} = (0, 1) = i. $$

  • $\begingroup$ Thank you sir for your answer. I want to ask why do e care that much about continuity? all of this confusion will disappear if we just made $i:= \sqrt{-1}$, $\endgroup$
    – pie
    Commented Mar 5 at 20:47
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    $\begingroup$ We do define $\sqrt{-1} = i$ (or, some of us do), but that definition doesn't mean we can manipulate complex square roots the same way we manipulate real roots. Particularly, $\sqrt{ab} = \sqrt{a} \sqrt{b}$ is simply not an identity for complex numbers, no matter how we define the square root. :) <> That is, I think it's not a matter of caring about (dis)continuity, it's a matter of ... topology being inextricably involved with the complex logarithm and root functions. $\endgroup$ Commented Mar 5 at 23:05

If $r$ is a non-negative real number, then $-r\notin\Bbb C\setminus(-\infty,0]$, and therefore $f$ is undefined at $-r$.


The constructions of complex number that you must have seen rely on people's intuitions and experience with arithmetic, in order to avoid certain formalities that are deemed too confusing or too demotivating for most people. However, if one is very careful, one discovers that they do not hold water, just like you did: how can we define $\sqrt{\phantom{x}}$ using $i$ and the same time define $i$ using $\sqrt{\phantom{x}}$? If that's not what we do, then what is $i$? (Another dilemma that sometimes arises is: thee are two square roots of $-1$, how do we know which one is $i$?)

The answer thar resolves the problems was already given by infinitylord, but they perhaps did not put the right emphasis on it to get through to you. So let me spell out the logical steps of the construction of $\mathbb{C}$.

Before we do that, you must forget all the talk about $i = \sqrt{-1}$ and such. All of that is just motivation and is not part of an actual, precise construction (even though many pre-university texts will pretend it is).

So, with a blank mind:

  1. We define $\mathbb{C} = \mathbb{R} \times \mathbb{R}$. That is, a complex number is a pair of real numbers.
  2. We define operations on complex numbers:
    • $(x,y) +_\mathbb{C} (u,v) = (x + u, y + v)$
    • $(x,y) -_\mathbb{C} (u,v) = (x - u, y - v)$
    • $(x,y) \cdot_\mathbb{C} (u,v) = (x \cdot u - y \cdot v, x \cdot v + y \cdot u)$
  3. We prove that with these operations $\mathbb{C}$ is a field.

Observe that we never mentioned square roots or $i$ to construct $\mathbb{C}$.

Now we define $i = (0,1)$. Still, we have not mentioned any square roots.

For historical reasons people use the following notational conventions:

  1. For any real number $x \in \mathbb{R}$, we are allowed to write $x$ instead of $(x,0)$. The reader is supposed to guess whether $x$ is to be read as a real or as $(x, 0)$.

  2. We write $+$, $-$, $\cdot$ for operations on reals and for operations on complex numbers, even thought they are different things. The reader is supposed to guess what is going on.

For example, if I write $2 + 3 \cdot i$ you are supposed to guess that this means $$(2,0) +_\mathbb{C} (3,0) \cdot_\mathbb{C} (0,1),$$ after which you can compute easily that the complex number in question is $(2, 3)$.

For any real numbers $x$ and $y$ we have $(x, y) = (x, 0) +_\mathbb{C} (y, 0) \cdot_\mathbb{C} (0,1)$, which we could write using the above conventions as $(x,y) = x + y \cdot i$. For historical reasons nobody writes $(x, y)$, and everyone prefers $x + y \cdot i$. But keep in mind that this is only possible because of the above conventions.

Textbooks skip over this explanation and just tell you that "every complex number has a real component $x$ and an imaginary component $y$, and can therefore be written as $x + i y$". The problem with this is that it confuses those who are smart enough (and the rest could not care less).

Now the definition of $\sqrt{\phantom{x}}$ is not mysterious at all. We defined $i$ without any mention of square roots, so we are free to mention $i$ when we define square roots.

I hope this clears up the confusion.

Addendum: Apparently confusion has not been cleared up. As my last attempt, I will define square root of a complex number. While doing so, we can use square roots of non-negative reals (but not of negative reals). Define the map $s : \mathbb{C} \to \mathbb{C}$ by the rule (we continue to write complex numbers as pairs of reals): $$ s(x,y) = \begin{cases} \left(\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}},-\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\right)& \text{if $y < 0$}\\ (0, \sqrt{-x})& \text{if $y = 0$ and $x < 0$}\\ (\sqrt{x}, 0)& \text{if $y = 0$ and $x \geq 0$}\\ \left(\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}},\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\right)& \text{if $y > 0$} \end{cases} $$ Note that all square roots in the definition of $s(x,y)$ are applied to non-negative real numbers (in particular $\sqrt{x^2+y^2}-x \geq 0$).

Theorem: For all $(x,y) \in \mathbb{C}$ we have $s(x,y) \cdot s(x,y) = (x,y)$.

Proof. This requires some calculation that I am not going to write down (I checked with Mathematica). Let us just see what happens when $x < 0$ and $y = 0$ (again, note that the numbers appearing under square roots are all non-negative reals):

\begin{align*} s(x,0) \cdot s(x,0) &= (0,\sqrt{-x}) \cdot (0,\sqrt{-x}) \\ &= (0 - \sqrt{-x} \cdot \sqrt{-x}, 0 \cdot \sqrt{-x} + \sqrt{-x} \cdot 0) \\ &= (0 - (-x), 0) \\ &= (x, 0). \end{align*} as required. QED.

We define the square root of a complex number to be $\sqrt{(x,y)} = s(x,y)$.

We calculate that $\sqrt{-1} = \sqrt{(-1,0)} = s(-1,0) = (0,\sqrt{-(-1)}) = (0, \sqrt{1}) = (0, 1) = i$.

  • $\begingroup$ The problem with this construction of $\mathbb{C}$ that it avoids the square roots of negatives entirely and this means we cannot use $i =\sqrt{-1}$ if we need to. for example if for some reason I need to break $x^2+1$ to $(x-i)(x+i)$ or If need to take an annoying negative number under the sqrt function in some integral now I can't do that. $\endgroup$
    – pie
    Commented Mar 15 at 16:58
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    $\begingroup$ Of course you can! Compute $(x - i) (x + i)$ in the given definition of $\mathbb{C}$, where $i = (0,1)$, and you will get $x^2 + (1,0)$. You can also easily compute that $(0,1)^2 = (-1,0)$. $\endgroup$ Commented Mar 15 at 22:44
  • $\begingroup$ So I can still take square root of negatives ? so why do we exclude that from the definition of the principal branch of $\sqrt$ ? I mean for example solution of $x^3- 15x-4 $ is $x=\sqrt[3]{2-\sqrt{-121}}+\sqrt[3]{2+\sqrt{-121}}$ so I can just use $i =\sqrt{-1}$ $\endgroup$
    – pie
    Commented Mar 15 at 22:57
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    $\begingroup$ You cannot just write $\sqrt{\mathrm{banana}}$ and pretend that's meaningful. Likewise, writing down $\sqrt{-1}$ without explaining what that means does not make it a number. The construction I described explains how to correctly define complex numbers. After we have done that, we can also define square root of a complex number. $\endgroup$ Commented Mar 15 at 22:59
  • $\begingroup$ Would you like me to define square roots of complex numbers explicitly? $\endgroup$ Commented Mar 15 at 23:00

Understanding the Square Root Function in Complex Analysis

The square root function on the real numbers is straightforward: for $ x \in \mathbb{R}^+$, $\sqrt{x}$ is the positive number which, when squared, returns $x$. Extending this function to the complex plane introduces several complications and requires careful definition to avoid ambiguity, primarily because every non-zero complex number has two square roots.

The principal branch of the square root function, $\sqrt{z}$, is defined to be single-valued. By restricting the argument of $z$ to $(-\pi, \pi]$ and choosing the square root with a non-negative imaginary part, we uniquely define $\sqrt{z}$ for all $z$ not on the non-positive real axis.

Understanding $\sqrt{-r}$ with the Principal Square Root

For any non-negative real number $r$, $\sqrt{-r}$ is not within the realm of real numbers, since no real number squared gives a negative result. We must extend to the complex plane, and by the definition you've encountered in Gamelin's book, we evaluate the principal square root of a negative real number as follows:

  • We have $z=-r$
  • The principle modulus of $z = -r$ is $|z| = r$.
  • The principle argument of $z = -r$ is $Arg(-r) = \pi$ because it lies on the negative real axis.
  • Using the definition, the principle square root is $\sqrt{z} = |z|^{1/2} e^{i\frac{Arg(z)}{2}}$.
  • Since $|z| = r$ and $Arg(z) = Arg(-r) = \pi$, we get $\sqrt{z} = \sqrt{r} e^{i \frac{\pi}{2}} = \sqrt{r} \cdot i = i\sqrt{r}$.

The Role of $i$ and the Definition of $i$

The symbol $i$ is introduced as the imaginary unit in the complex plane. It indeed is defined as a number that satisfies $i^2 = -1$. However, this is not equivalent to defining $i$ as the square root of $-1$, because there are two complex numbers that square to $-1$: $i$ and $-i$.

When mathematicians say $i = \sqrt{-1}$, it's shorthand for the more precise statement that $i$ is the number that, by convention, is as the principal square root of $-1$, but this is more of a notational convenience than a definition derived from the square root function.

Why Continuity Matters

We care about continuity because it's a fundamental concept in analysis that allows us to apply calculus (ex. differentiation and integration) and other powerful mathematical tools. Continuous functions allow us to understand their behavior both locally and globally. Discontinuities, while interesting and important in their own right, often require more careful handling to understand the behavior of functions around them. If a function isn't continuous, we lose the ability to use these tools or must apply them with great care.

The principal branch of the square root function is defined in such a way as to make it continuous on its domain, meaning that as one moves smoothly around the complex plane, the value of the square root changes smoothly too, without any jumps or breaks. For this reason we avoid defining $\sqrt{z}$ for $ z \in \mathbb{R}^-$ directly.

The Practical Use of $i = \sqrt{-1}$

While it's true that in many practical applications $i$ is used as if it were simply $\sqrt{-1}$, this is a convenience that sidesteps the underlying complexities. This simplification is often safe in contexts that you mentioned such as high school algebra, some engineering applications, and certain physics problems because, in those contexts, the exact nature of $i$ as a principal square root is not as crucial.

However, in higher mathematics, especially in fields like complex analysis, we need to be precise about such definitions. Ambiguities can lead to errors and misinterpretations, which is why the formalism of defining $i$ as a unit that squares to $-1$, and carefully defining the principal branch of the square root function is necessary.


Although $i$ is commonly introduced as $\sqrt{-1}$ in many contexts, in the context of complex analysis, we introduce it as a number that squares to $-1$ and then define the principal branch of the square root function to maintain continuity and unambiguity. Your graduate friends likely work with this simplified definition that's established by convention because it is practical for their work, but may lack the rigor that complex analysis demands.


One way in which $i$ can be defined without referring to the square root function or continuity at all comes from polynomial algebra. This approach requires knowledge of concepts that are typically introduced in a first course in abstract algebra, but I'll try to keep my answer somewhat self-contained. However, I will be assuming that you are comfortable with equivalence relations and equivalence classes.

We start with polynomials with real coefficients, denoted $\mathbb R[x]$. We are interested in the polynomial $x^2 + 1$, in particular we are interested in solutions to $x^2 + 1 = 0$.

In abstract algebra this is expressed in the language of ideals. An ideal of $\mathbb R[x]$ generated by a polynomial $f$, denoted $(f)$, is the subset of polynomials that have $f$ as a factor. We focus on the ideal $(x^2+1)$.

We can define an equivalence relation $\equiv$ on $\mathbb R[x]$ by declaring $f \equiv g$ iff $f - g$ has $x^2+1$ as a factor (prove as an exercise). The quotient space of $\mathbb R[x]$ with respect to the equivalence class $\equiv$ is denoted by $\mathbb R[x]/(x^2+1)$. Lets denote the equivalence class in $\mathbb R[x]/(x^2+1)$ represented by a polynomial $f\in\mathbb R[x]$ with $[f]$.

Note that by our definition, $[x^2+1]\equiv [0]$ as obviously $x^2+1-0$ has $x^2+1$ as a factor.

Now the important thing is that addition and multiplication of equivalence classes in $\mathbb R[x]/(x^2+1)$ are well defined, i.e. we define $[f]+[g]\equiv[f+g]$ and $[f][g] \equiv[fg]$, and this does not depend on the choice of representative (prove as an exercise). In fact, with this definition of multiplication and addition the quotient space $\mathbb R[x]/(x^2+1)$ is a field, with $[0]$ as the neutral element for addition and $[1]$ for multiplication.

Consider now the equivalence class $[x]$. We have $[x]^2 \equiv [x^2]$ and clearly $[x^2] \equiv [-1]$.

Let's give a special name to this class $[x]$, let's call it $\iota$. Then $\iota^2 \equiv [-1]$.

If we want to abuse notation some more, we can use the $=$ symbol for our equivalence relation, so now $\iota^2 = [-1]$. Moreover, if we keep a mental note that we're talking about equivalence classes, we can drop the $[ \cdot ]$ notation and have $\iota^2 = - 1$.

All this abuse is made formal with the concepts of ring, quotient ring, maximal ideal and ring isomorphism. It can be shown that the quotient ring $\mathbb R[x]/(x^2+1)$ is isomorphic to $\mathbb C$, where $\mathbb C$ is defined as in the other answers.


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