The Square Root Function: From Real Numbers to Complex Analysis

Assumption is the mother of all screw-ups, Anonymous.

Introduction

The square root function is a fundamental mathematical function written as $f(x) = \sqrt{x}$ with wide-ranging applications in mathematics, science, and engineering.

It gives or returns the non-negative number that, when multiplied by itself, equal the input x. For example, $\sqrt{16}$ = 4 because 4 × 4 = 16, $\sqrt{49}$ = 47 because 7 × 7 = 49, $\sqrt{64}$ = 8 because 8 × 8 = 64.

Basic Properties

• Domain: The domain of the square root function is $[0, \infty)$ — the non-negative real numbers. In the realm of real numbers $\mathbb{R}$, you can’t take the square root of a negative number (it is undefined).
• Range: The set of all possible outputs is also the non-negative real numbers $[0, \infty)$.
• Increasing Function: The square root function is stricly increasing: if $0 \leq a < b$, then $\sqrt{a} < \sqrt{b}$, meaning that as the input gets larger, the output also gets larger.
• The graph is continuous for all $x \geq 0$ and smooth (infinitely differentiable) for all x > 0. It starts at the origin (0, 0) and curves upward and to the right. As x approaches infinity, the function grows without bound, but the rate of growth slows down. It is a concave down curve because its second derivative is negative (since $f''(x) < 0$ for $x > 0$).
• The square root function f(x) = $\sqrt{x}$ is the inverse of the squaring function $g(x) = x^2$, but only when the domain of g(x) is restricted to non-negative numbers ($x \geq 0$). Graphically, the curve $y = \sqrt{x}$ is the reflection of $y = x^2$ (for $x \geq 0$) across the line $y = x$.

Fundamental Identities

• Basic Identities: (1) Inverse Property, $(\sqrt{x})^2 = x$ for $x \geq 0$. (2) Absolute Value Identity, $\sqrt{x^2} = |x|$ for all $x \in \mathbb{R}$.
• Algebraic Properties: (1) Product Rule, $\sqrt{ab} = \sqrt{a}·\sqrt{b}$ for $a, b \geq 0$; (2) Quotient Rule, $\sqrt{a/b} = \sqrt{a}/\sqrt{b}$ for $a \geq 0$, $b > 0$.

Calculus

• The derivate of the square root function $\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}$ for $x > 0$. Note that $f'(x) \to \infty$ as $x \to 0^+$: the graph has a vertical tangent at the origin.
• Integral: The integral of $\sqrt{x}$ is $\frac{2}{3}x^{3/2} + C$.

The Complex Square Root Function

A complex number is expressed as $z = a + bi$, where $a$ and $b$ are real numbers ($a, b \in \mathbb{R}$) and $i$ is the imaginary unit ($ i^2 = -1 $). The square root of a complex number $ z $ is any number $ w $ such that $ w^2 = z $.

Unlike real numbers, every non-zero complex number has exactly two distinct square roots, and these roots are negatives of each other. The number $0$ has the unique square root $0$. This is fundamentally different from the real case, where negative numbers have no real square roots at all.

Proof. Suppose $w \neq 0$. If $w^2 = z$ and $w'^2 = z$, then $w^2 - w'^2 = 0$, so $(w - w')(w + w') = 0$. Thus, either $w' = w$ or $w' = -w$. Since $w \neq 0$ (as $z \neq 0$), the two roots $w$ and $-w$ are distinct.

Polar-form derivation

The most straightforward and elegant method to compute complex square roots uses polar form. For any nonzero complex number z, we can write it as $z = re^{i\theta}$ where r = |z| > 0 is the modulus (a positive real number, distance from the origin) and $\theta = \arg(z)$ is the argument (angle from the positive real axis), typically taken in (-π, π] or [0, 2π).

Taking the square root means finding $w$ with $w^2 = z$. If $w = \rho e^{i\phi}$, then: $w^2 = \rho^2 e^{2i\phi} = re^{i\theta}.$

Comparing moduli and arguments: $\rho^2 = r \;\Longrightarrow\; \rho = \sqrt{r}, \qquad 2\phi = \theta + 2\pi k \;\Longrightarrow\; \phi = \frac{\theta}{2} + \pi k.$

For $k = 0$ and $k = 1$, we obtain the two distinct square roots: $\boxed{w_1 = \sqrt{r}\,e^{i\theta/2}, \qquad w_2 = \sqrt{r}\,e^{i(\theta/2 + \pi)} = -w_1.}$

In trigonometric form: $w_1 = \sqrt{r}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right), \qquad w_2 = \sqrt{r}\left(\cos\!\left(\frac{\theta}{2} + \pi\right) + i\sin\!\left(\frac{\theta}{2} + \pi\right)\right).$

Steps to Find Square Roots:

1. Compute the modulus of the roots: $ \sqrt{r} $.
2. Compute the arguments of the roots: $ \frac{\theta}{2} $ and $ \frac{\theta}{2} + \pi $.
3. The two square roots are. $\sqrt{r}e^{i\theta/2}, \sqrt{r}e^{i(\theta/2+\pi)}$
4. Equivalently, in trigonometric form: $w_1 = \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i\sin\left(\frac{\theta}{2}\right) \right), w_2 = \sqrt{r} \left( \cos\left(\frac{\theta}{2} + \pi\right) + i\sin\left(\frac{\theta}{2} + \pi\right) \right) = -w_1$

Why Two Roots?

Since angles are defined modulo $2\pi$, halving $\theta$ produces two distinct values modulo $2\pi$: $\frac{\theta}{2} \quad \text{and} \quad \frac{\theta}{2} + \pi.$

These differ by $\pi$ radians — exactly half a full rotation — placing them on diametrically opposite sides of the origin in the complex plane. For $k = 2$, we get $\frac{\theta}{2} + 2\pi$, which is the same angle as $\frac{\theta}{2}$ (mod $2\pi$), so no new root appears.

In other words, the modulus of the root is the square root of the original modulus. The angle of the root is half the original angle. The second root lies diametrically opposite the first root - the two roots are antipodal separated by $\pi$ radians on a circle of radius $\sqrt{r}$.

Rectangular-form formula

For $z = a + bi$ with $b \neq 0$, we can express the square roots directly in terms of $a$, $b$, and $|z|$.

If z = a + bi with z ≠ 0, assume the square roots of z to be w = u + iv, that is $\sqrt{a + bi} = u + iv$

Now, squaring both side of the equation, we have: $a + bi = u^2 -v^2 + 2iuv$. Comparing real and imaginary parts of the above equation, we have: $a = u^2 -v^2, b = 2uv$

We also know that $(u^2 + v^2)^2 = (u^2 - v^2)^2 + 4u^2v^2 = a^2 + b^2 \leadsto u^2 + v^2 = \sqrt{a^2 + b^2 }$ taking the positive root since $u^2 + v^2 > 0$.

Now, we have the system: $u^2 + v^2 = |z|, \qquad u^2 - v^2 = a.$

Adding and subtracting: $u^2 = \frac{|z| + a}{2}, \qquad v^2 = \frac{|z| - a}{2}.$

Both quantities are non-negative: $|z| + a \geq 0$ because $|z| \geq |a| \geq -a$, and $|z| - a \geq 0$ because $|z| \geq |a| \geq a$.

The sign relationship comes from $b = 2uv$: $u$ and $v$ have the same sign when $b > 0$, and opposite signs when $b < 0$.

Using the sign function $\operatorname{sgn}(b)$: $\boxed{\sqrt{a + bi} = \pm\left(\sqrt{\frac{|z| + a}{2}} + i\,\operatorname{sgn}(b)\sqrt{\frac{|z| - a}{2}}\right)}$ where $|z| = \sqrt{a^2 + b^2}$ and the $\pm$ gives both roots.

Special case $b = 0$. If $z = a \in \mathbb{R}$: (i) $a \geq 0$: $\sqrt{a} = \pm\sqrt{a}$ (the familiar real square roots). (ii) $a < 0$: $\sqrt{a} = \pm i\sqrt{|a|}$.

Examples

1. Square Root of -1, $z = -1 = 1 \cdot e^{i\pi}$. Roots: $w_1 = \sqrt{1} e^{i\pi/2} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i, w_2 = \sqrt{1}\,e^{i(\pi/2 + \pi)} = \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} = -i.$
2. Square Root of 1 + i, Polar method, $r = \sqrt{1^2 + 1^2} = \sqrt{2}, \theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}$. Roots: $w_1 = \sqrt[4]{2} e^{i\pi/8} = \sqrt[4]{2} \left( \cos\left(\frac{\pi}{8}\right) + i\sin\left(\frac{\pi}{8}\right) \right), w_2 = \sqrt[4]{2} e^{i({\pi/8 + \pi})} = \sqrt[4]{2} \left( \cos\left(\frac{9\pi}{8}\right) + i\sin\left(\frac{9\pi}{8}\right) \right) \text{ where } \frac{9\pi}{8} = \frac{\pi}{8} + \pi$
3. Square Root of 1 + i. $a = 1$, $b = 1$, $|z| = \sqrt{2}$. Rectangular method $u = \sqrt{\frac{\sqrt{2} + 1}{2}}, \qquad v = \operatorname{sgn}(1)\sqrt{\frac{\sqrt{2} - 1}{2}} = \sqrt{\frac{\sqrt{2} - 1}{2}}.$
   $\sqrt{1 + i} = \pm\left(\sqrt{\frac{\sqrt{2} + 1}{2}} + i\sqrt{\frac{\sqrt{2} - 1}{2}}\right).$

Branches and the Principal Square Root

Unlike real numbers, every non-zero complex number has exactly two distinct square roots. The challenge is defining a consistent function, $\sqrt{z}$, that selects one of these two values.

This selection cannot be made continuously on all of $\mathbb{C} \setminus \{0\}$: Start at $z = 1$ with the square root $w = 1$. Traverse the unit circle $z = e^{i\theta}$ as $\theta$ increases from $0$ to $2\pi$: $w(\theta) = e^{i\theta/2}.$

(i) At $\theta = 0$: $w = e^0 = 1$. (ii) At $\theta = 2\pi$ (back to $z = 1$): $w = e^{i\pi} = -1$. The function has not returned to its starting value — it has jumped to the other root. This demonstrates that no continuous single-valued square root exists on $\mathbb{C} \setminus \{0\}$. Traversing the loop a second time returns $w$ to $+1$: the square root has a monodromy of order 2 (if you go around a loop twice, the sign flips twice: $+1\rightarrow -1\rightarrow +1$).

The multi-valuedness of the square root is inherited from the logarithm. The most fundamental definition of $z^{1/2}$ is: $\boxed{z^{1/2} = e^{\frac{1}{2}\log(z)}}$.

Since $\log(z) = \ln r + i(\theta + 2\pi k)$ for $k \in \mathbb{Z}$: $z^{1/2} = e^{\frac{1}{2}(\ln r + i(\theta + 2\pi k))} = \sqrt{r}\,e^{i\theta/2}\,e^{i\pi k} = \boxed{\sqrt{r} e^{i\theta/2} (-1)^k.}$

$e^{i\pi n} = (-1)^n$.

• Even $k$ ($k = 0, 2, 4, \ldots$): $(-1)^k = 1$, yielding $w_1 = \sqrt{r}\,e^{i\theta/2}$.
• Odd $k$ ($k = 1, 3, 5, \ldots$): $(-1)^k = -1$, yielding $w_2 = -\sqrt{r}\,e^{i\theta/2}$.

A branch cut is a curve (or line) in the complex plane that we agree not to cross. Thus, the infinitely many values of $\log(z)$ collapse to exactly two values for $z^{1/2}$. Choosing a branch of the logarithm determines a branch of the square root.

Definition. The principal square root is defined by: $\boxed{\sqrt{z} = e^{\frac{1}{2}\operatorname{Log}(z)} = \sqrt{|z|} e^{i\operatorname{Arg}(z)/2}},$ where $\operatorname{Log}(z) = \ln|z| + i\operatorname{Arg}(z)$ is the principal logarithm with $\operatorname{Arg}(z) \in (-\pi, \pi]$.

Domain. $\mathbb{C} \setminus (-\infty, 0]$ — the complex plane with the negative real axis (and the origin) removed. This is the branch cut, inherited from the principal logarithm. Those two previous values correspond to the two branches of the square root function:

• The principal branch cut is the negative real axis: $f_+: \Complex \setminus (-∞, 0] \to \Complex, f_+(z) = \sqrt{r} \cdot e^{i\theta/2}$ where $\theta(z) \in (-\pi, \pi], \theta(z) \in arg(z)$. Define $\boxed {\sqrt{z} = e^{\frac{1}{2}Log(z)} = \sqrt{|z|}e^{i\theta/2}}$ where $Log(z)$ is the principal logarithm.
• Secondary branch is on the same domain, $f_-: \Complex \setminus (-∞, 0] \to \Complex, f_-(z) = \sqrt{r} \cdot e^{i\theta/2+i\pi}$ where $\theta(z) \in (\pi, 2\pi], \theta(z) \in arg(z)$

Properties of the principal branch $f_+(z) = \sqrt{z}$:

• Single-valued: assigns exactly one value to each $z$ in the domain. By restricting $Arg(z) \in (-\pi, \pi]$, we select exactly one of the two possible square roots for each $z \ne 0$.
• Holomorphic: $f_{+}(z) = \sqrt{z}$ is complex-differentiable at every point of $\mathbb{C} \setminus (-\infty, 0]$ with derivative $\frac{d}{dz}\sqrt{z} = \frac{1}{2\sqrt{z}}$.
• Right half-plane image: Since $\operatorname{Arg}(z) \in (-\pi, \pi)$ on the slit domain, we have $\operatorname{Arg}(\sqrt{z}) = \operatorname{Arg}(z)/2 \in (-\pi/2, \pi/2)$. Therefore $\operatorname{Re}(\sqrt{z})=\sqrt{r}cos(\operatorname{Arg}(z)/2) > 0$ — the principal square root always has positive real part.

  By restricting the argument of z to $(-\pi, \pi)$, we force the argument of $\sqrt{z}$ into $(-\pi /2,\pi /2)$, which is a region that does not wrap around the origin. No looping, no branch switching -the principal square root is single‑valued on the slit plane.

• Continuity at the positive real axis: for $x > 0$, $Arg(x) = 0, \sqrt{x}=\sqrt{x}e^{i\cdot 0} = \sqrt{x}$ equals the familiar positive real square root. The principal branch agrees with the familiar positive real square root on $\mathbb{R}^+$.
• Discontinuity across the cut: approaching the negative real axis from above ($z = -r + i\epsilon$, $\epsilon \to 0^+$) gives $\operatorname{Arg}(z) \to \pi^-$ and $\sqrt{z} \to \sqrt{r}e^{i\pi/2} = i\sqrt{r}$. From below ($z = -r - i\epsilon$, $\epsilon \to 0^+$), $\operatorname{Arg}(z) \to -\pi^+$ and $\sqrt{z} \to \sqrt{r}e^{-i\pi/2} = -i\sqrt{r}$. The function jumps by $2i\sqrt{r}$ across the cut.

Differentiation of the Square Root

1. The principal branch cut is the negative real axis: $f_+: \Complex \setminus (-∞, 0] \to \Complex, f_+(z) = \sqrt{r} \cdot e^{i\theta/2}$ where $\theta(z) \in (-\pi, \pi], \theta(z) \in arg(z)$. Define $\boxed {\sqrt{z} = e^{\frac{1}{2}Log(z)} = \sqrt{|z|}e^{i\theta/2}}$ where $Log(z)$ is the principal logarithm. Thus, $\sqrt{z}$ is analytic on the domain $\Complex \setminus (-∞, 0]$.
2. Derivative via implicit differentiation. By definition, f(z) satisfies $f_+(z)^2 = z, \forall z \in \Complex \setminus (-∞, 0]$. Differentiating both sides using the chain rule, we get: $2f_+(z) \cdot f_+'(z) = 1 \quad \Longrightarrow[\text{Solving for f'(z) and substituting back } f(z) = \sqrt{z}] f_+'(z) = \frac{1}{2f_+(z)} = \frac{1}{2\sqrt{z}}.$
3. This is valid on $\mathbb{C} \setminus (-\infty, 0]$ (and $z \neq 0$ because the square root itself is not analytic at z = 0). The formula is identical to the real derivative $\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}$.
4. Derivative using the logarithmic definition. Using the explicit definition of the principal branch, $\sqrt{z} = e^{\frac{1}{2}\operatorname{Log}(z)}$, we differentiate directly, $\frac{d}{dz}\sqrt{z} = \frac{d}{dz}e^{\frac{1}{2}\operatorname{Log}(z)} = e^{\frac{1}{2}\operatorname{Log}(z)} \cdot \frac{1}{2z} = \frac{\sqrt{z}}{2z} = \frac{1}{2\sqrt{z}},$ the same result as before.

   Since $z \ne 0$, we may rewrite $\frac{\sqrt{z}}{2z}$ as $\frac{1}{2\sqrt{z}}$ because $\sqrt{z} \cdot \sqrt{z}$.

5. The derivative exists and is analytic everywhere on the chosen branch’s domain, i.e., the complex plane minus the branch cut.

Connection to the Complex Logarithm

A more fundamental way to define the square root is via the complex exponential and logarithm functions: $z^{1/2} = e^{\frac{1}{2}log(z)}$

The multi-valued nature of the square root comes directly from the multi-valued nature of the logarithm: log(z) = ln(r) + i(θ + 2πk).

Substituting this into the formula gives: $z^{1/2} = e^{\frac{1}{2}log(z)} = e^{\frac{1}{2} \left( \ln r + i(\theta + 2\pi k) \right)} = \sqrt{r}·e^{i\frac{\theta + 2\pi k}{2}} = \sqrt{r}·e^{i\frac{\theta}{2}}·e^{i\pi k} = \sqrt{r}·e^{i\frac{\theta}{2}}·(-1)^k$.

This shows that an even k (like k = 0, 2,…) results in one root ($e^{i\pi k} = 1$), while an odd k (like k=1,3,…) results in the other ($e^{i\pi k} = - 1$). Choosing the principal branch of the logarithm corresponds to choosing the principal branch of the square root.

Complex exponentiation

The core issue is that the complex logarithm, log(z), is a multi-valued function. For any complex number $z = re^{i\theta}$, its logarithm is log(z) = ln(r) + i(θ + 2πk) where k can be any integer (k ∈ ℤ). Each value of k produces a different value for the logarithm.

When we define $z^α := e^{α·log(z)}$. We define an analytic branch of $z^α$ by choosing a branch of Log(z), this multi-valuedness carries over: $z^α = e^{α·(ln(r) + i(θ + 2πk))} = e^{α·(ln(r) + iθ)}e^{α·i2πk}$

Let’s analyze the two parts of this product:

1. $e^{α·(ln(r) + iθ)}$: This part depends only on a single choice of θ. It forms the principal value if we choose the principal argument θ ∈ (−π, π].
2. $e^{α·i2πk}$: This is the crucial term, the multi-valued factor. It’s a multiplier that generates all the possible values of $z^α$.

The number of distinct values depends on α:

1. If α is an integer: Let α = n. The multiplier becomes $e^{i2πnk} = (e^{i2π})^{nk} = 1^{nk} = 1.$ There is only one value, as expected (z², z³, etc., are unique).
2. If α is a rational number: Let α = p/q. The multiplier is $e^{i2π·\frac{p}{q}·k}$. As $k$ varies, this term cycles through a finite set of values. It produces exactly $q$ distinct values (corresponding to $k = 0, 1, 2, ..., q-1$). This explains why there are $q$ distinct $q$-th roots ($z^{1/q}$) and, more generally, $q$ distinct values for $z^{p/q}$.
3. If α is an irrational or complex number: The multiplier $e^{i2πnk}$ produces an infinite number of distinct values, one for each integer k.

To create a proper, single-valued, and analytic (differentiable) function from $z^α$, we must make a consistent choice. By choosing a branch for log(z), we are essentially fixing the value of k (usually k = 0) and restricting the domain of the argument θ. The principal branch of zᵅ is defined by using the principal branch of the logarithm, Log(z) = ln∣z∣ + iArg(z), where Arg(z) ∈ (−π, π].

This gives the principal Value of $z^α$: $z^α = e^{α·Log(z)}$ where Arg(z) is the principal argument in (−π,π]. A key feature of the principal root is that its real part is always non-negative. It maps the entire complex plane (minus the branch cut) to the right half-plane.

Properties of the Principal Branch:

• Single-Valued: It yields one specific result for a given $z$.
• Analytic: It is a differentiable function everywhere except on its branch cut.
• Branch Cut: The function is discontinuous and non-analytic along the branch cut inherited from $\operatorname{Log}(z)$, which is the negative real axis (${z \in \mathbb{C} : \operatorname{Im}(z) = 0, \operatorname{Re}(z) \leq 0}$). This is because $\operatorname{Arg}(z)$ jumps from $π$ to $-π$ across this ray.

Example:

• $i^i = e^{i·Log(i)} =[Log(i) = ln(1) + iπ/2 = iπ/2] e^{i·(i\frac{\pi}{2})} = e^{-\frac{\pi}{2}}$ So, the principal value of $i^i$ is the real number $e^{-\frac{\pi}{2}}≈0.20788.$

The full-multivalued set is iⁱ = {$e^{-\frac{\pi}{2}}·e^{i·i2πk} = e^{-\frac{\pi}{2}-2πk} | k \in \mathbb{Z}$}

• $(-1)^i = e^{i·Log(-1)} =[Log(-1) = ln(1) + iπ = + iπ] e^{i·(i\pi)} = e^{-\pi}$ So, the principal value of $(-1)^i$ is the real number $e^{-\pi}≈0.0432.$

The full-multivalued set is (-1)ⁱ = {$e^{-\pi}·e^{i·i2πk} = e^{-(\pi + 2\pi k)} = e^{-(\pi + 2\pi k)} = e^{-(2k+1)\pi} | k \in \mathbb{Z}$}

Derivatives

• If α is a constant and a fixed branch Log, $f(z) = z^\alpha = e^{\alpha·Log(z)}, f'(z) = e^{\alpha·Log(z)}·\alpha·(Log(z))^' = z^\alpha·\alpha·\frac{1}{z} = \alpha z^{\alpha -1}$ valid on the branch domain (away from the cut and z = 0).
• If α = α(z) varies with z, $\frac{d}{dz}(z^\alpha(z)) = e^{\alpha(z)·Log(z)}·(\alpha^'(z)Log(z) + \frac{\alpha(z)}{z}) = z^\alpha(z)·(\alpha^'(z)Log(z) + \frac{\alpha(z)}{z})$.