Complex Square Root, Exponentiation, and the Quadratic Formula

Beware that, when fighting monsters, you yourself do not become a monster… for when you gaze long into the abyss. The abyss gazes also into you, Friedrich W. Nietzsche

Complex square root

For any nonzero complex number $z = x + iy = re^{i\theta}$ where $r = |z|$ and $\theta = \text{Arg}(z)$, the square roots are the two numbers $w$ satisfying $w^2 = z$.

Using the general $n$-th root formula with $n = 2$: $w = \sqrt{|z|} \cdot e^{i\frac{\theta + 2\pi k}{2}}, \quad k = 0, 1$. This gives exactly two square roots:

• Principal Square Root ($k = 0$). $\boxed{\sqrt{z} = \sqrt{|z|} \cdot e^{i\theta/2} = \sqrt{|z|}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)}$ where $u = \sqrt{|z|}\cos\frac{\theta}{2}, \quad v = \sqrt{|z|}\sin\frac{\theta}{2}$
• The Other Root ($k = 1$). $-\sqrt{z} = \sqrt{|z|} \cdot e^{i(\theta/2 + \pi)} = \sqrt{|z|}\left(\cos\left(\frac{\theta}{2} + \pi\right) + i\sin\left(\frac{\theta}{2} + \pi\right)\right)$. Since $\cos(\alpha + \pi) = -\cos\alpha$ and $\sin(\alpha + \pi) = -\sin\alpha$: $\boxed{-\sqrt{z} = -\sqrt{|z|}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)}$

  The two square roots are negatives of each other, just like in the real case.

The formula $w = \sqrt{z}e^{\frac{i\theta}{2}}$ has a beautiful geometric meaning:

• Modulus: The distance from the origin is square-rooted. If z is at distance r, w is at distance $\sqrt{r}$. This “compresses” the plane: the exterior of the unit circle (r > 1) moves closer to the circle, and the interior (r < 1) moves closer to the origin.
• Argument: The angle is halved. If z is at angle θ, w is at angle θ/2.
• Two roots: Because $e^{i(\frac{\theta}{2}+\pi)} = -e^{\frac{i\theta}{2}}$, the two roots are always diametrically opposite each other on a circle of radius $\sqrt{r}$.
• Example: For z = i (r = 1, $\theta = \frac{\pi}{2}$), $w_1 = e^{\frac{i\pi}{4}} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}, w_2 = e^{\frac{i5\pi}{4}} = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$

Derivation of the Cartesian Formula

Let w = u + iv such that $w^2 = z = x + iy$. Expanding: $(u + iv)^2 = u^2 -v^2 + 2iuv = x + iy$. This gives a system of two equations: (i) $u^2 -v^2 = x$ (real part); (ii) 2uv = y (imaginary part).

Besides, $r = |z| = \sqrt{x^2 + y^2} \implies r^2 = x^2 + y^2$. Crucially, $(u^2 + v^2)^2 = (u^2 - v^2)^2 + (2uv)^2 =[(i),(ii)] x^2 - y^2 = r^2 \implies[u^2 + v^2 \ge 0] u^2 + v^2 = r$ (iii).

(i) + (iii): $2u^2 = r + x \implies u = \pm \sqrt{\frac{r+x}{2}}$
(iii) - (i): $2v^2 = r - x \implies v = \pm \sqrt{\frac{r-x}{2}}$

From equation (ii), 2uv = y. The signs of u and v must be chosen so that their product has the same sign as y. If y > 0, u and v have the same sign. If y < 0, u and v have opposite signs. This is compactly written using the sign function: $v= sgn(y)\cdot \sqrt{\frac{r-x}{2}}$ (assuming we take the positive root for u).

For $z = x + iy$ with $r = |z| = \sqrt{x^2 + y^2}$: $\boxed{\sqrt{z} = \pm\left(\sqrt{\frac{r + x}{2}} + i \cdot \text{sgn}(y) \cdot \sqrt{\frac{r - x}{2}}\right)}$ where $\text{sgn}(y)$ is the sign of $y$ (with $\text{sgn}(0) = 1$ by convention).

Special Case (y = 0). If z is real, then y = 0. (ii) becomes 2uv = 0 $\implies u = 0 \text{ or } v = 0$.

1. If x > 0 (Positive Real Axis). We are looking for the square root of a positive number. We know from real arithmetic that the roots are real. Therefore, the imaginary part v must be 0. $r = |z| = \sqrt{x^2+0^2} = x$, $\sqrt{z} = \pm\left(\sqrt{\frac{r + x}{2}} + i \cdot \text{sgn}(y) \cdot \sqrt{\frac{r - x}{2}}\right) =[r = x] \pm(\sqrt{x}+0i) = \pm\sqrt{x}$, two real numbers on the real axis, symmetric about the origin.
2. Case B: x < 0 (Negative Real Axis). We are looking for the square root of a negative number. We know from real arithmetic that the roots are imaginary. Therefore, the real part u must be zero. $r = |z| = \sqrt{x^2+0^2} = -x$, $\sqrt{z} = \pm\left(\sqrt{\frac{r + x}{2}} + i \cdot \text{sgn}(y) \cdot \sqrt{\frac{r - x}{2}}\right) =[r = -x] \pm(0+i\sqrt{-x}) = \pm i\sqrt{|x|}$, two purely imaginary numbers on the imaginary axis, symmetric about the origin.

Branch Cuts and Multi-Valuedness

The square root function is inherently two-valued. To make it single-valued, we:

1. Choose a branch cut: Typically the negative real axis $(-\infty, 0]$
2. Define the principal branch: $\sqrt{z}$ with $\text{Arg}(\sqrt{z}) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right]$

   $\text{Arg}(z) \in (-\pi, \pi] \implies \text{Arg}(\sqrt{z}) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right]$. This means the principal root always lies in the right half-plane.

3. Discontinuity: The principal square root is discontinuous across the branch cut (the negative real axis).
   If you approach the negative real axis from above ($\theta \to \pi$), the root approaches $i\sqrt{|x|}$. If you approach the negative real axis from below ($\theta \to -\pi$), the root approaches $-i\sqrt{|x|}$.

Properties of Square Root

Property	Formula
Product	$\sqrt{zw} = \pm\sqrt{z}\sqrt{w}$ True for the multi-valued function (i).
Quotient	$\sqrt{z/w} = \pm\sqrt{z}/\sqrt{w}$ True for the multi-valued function.
Conjugate	$\overline{\sqrt{z}} = \sqrt{\bar{z}}$ The principal branch is used on both sides and z not on the branch cut, (ii)
Modulus	$\vert \sqrt{z} \vert = \sqrt{\vert z \vert}$
Argument	$\arg(\sqrt{z}) = \frac{1}{2}\arg(z)$ (principal value)

(i) $\sqrt{zw} = \pm\sqrt{z}\sqrt{w}$ do not generally hold for the principal square root (or any single-valued branch). Let z = w = -1, $\sqrt{(-1)(-1)} = 1 \ne \sqrt{-1}\cdot \sqrt{-1} = -1$
(ii). If $z=r e^{i\theta}$ with $\theta \in (-\pi,\pi)$ (strict inequality avoids the cut at $\theta=\pi$), then $\overline{\sqrt{z}}=\overline{\sqrt{r} e^{i\theta/2}} = \sqrt{r}e^{-i\theta/2}.$
But $\overline{z}=r e^{-i\theta}$ and the principal argument of $\overline{z}$ is $-\theta$ (still in $(-\pi,\pi)$), so $\sqrt{\overline{z}}=\sqrt{r}e^{i(-\theta)/2}=\sqrt{r}e^{-i\theta/2}.$ Hence $\overline{\sqrt{z}}=\sqrt{\overline{z}}$ for all such z.

What goes wrong on the negative real axis?
Take z = -1. With the usual principal argument $Arg(-1)=\pi$, so $\sqrt{-1}= \sqrt{1}e^{i\pi/2}=i.$ Then $\overline{\sqrt{-1}}=-i$. But $\overline{-1}=-1$ and the principal square root of (-1) is again i. Thus $\overline{\sqrt{-1}}=-i\neq i=\sqrt{\overline{-1}}.$

Complex Exponentiation

For any complex number $z \neq 0$ and any complex power p, we can express the exponentiation as: $z^p = e^{log(z^p)} = e^{p\cdot log(z)}$ where $\mathbb{log(z)}$ is the (multi-valued) complex logarithm.

Recall the Complex Logarithm. The complex logarithm of a complex number $z \neq 0$ can be written as: $\mathbb{log(z)} = \mathbb{ln|z|} + i(Arg(z)+2πk), k \in ℤ$ where:

• ln denotes the natural logarithm of the magnitude (or modulus) of z.
• Arg(z) represents the principal argument of z, typically in the range (-π, π] or angle of z in the complex plane, which is the angle formed with the positive real axis.
• $2\pi k, k \in \mathbb{Z}$ accounts for the multivalued nature of the logarithm due to the periodicity of the argument.

Principal Logarithm: $\text{Log}(z) = \ln|z| + i\text{Arg}(z)$ (single-valued with branch cut).

Types of Complex Powers

Base $z$	Exponent $p$	Result	Multi-valued?
Complex	Integer	Complex	No
Complex	Rational $m/n$	Complex	$n$ values
Complex	Irrational real	Complex	Infinitely many
Complex	Complex	Complex	Infinitely many

Classical example. Calculate $i^i$ (z = i and p = i).

1. Apply the definition: $i^i = e^{i \cdot \log(i)}$.
2. Find $\log(i)$: $|i| = 1$, so $\ln|i| = 0$, $\text{Arg}(i) = \frac{\pi}{2}$. Then, $\log(i) = 0 + i\left(\frac{\pi}{2} + 2\pi k\right) = i\left(\frac{\pi}{2} + 2\pi k\right)$
3. Substitute: $i^i = e^{i \cdot i(\frac{\pi}{2} + 2\pi k)} = e^{i^2(\frac{\pi}{2} + 2\pi k)} = e^{-(\frac{\pi}{2} + 2\pi k)}$
4. Conclude: $\boxed{i^i = \left\{e^{-(\frac{\pi}{2} + 2\pi k)} : k \in \mathbb{Z}\right\}}$
5. Principal value ($k = 0$): $i^i = e^{-\pi/2} \approx 0.2079$. $i^i$ is real! All its values are positive real numbers.

General Example: $(1 + i)^{2 + 3i}$

1. Apply definition: $(1 + i)^{2 + 3i} = e^{(2 + 3i)\log(1 + i)}$.
2. Find $\log(1 + i)$: $|1 + i| = \sqrt{2}$ and $\text{Arg}(1 + i) = \frac{\pi}{4}$. $\log(1 + i) = \ln\sqrt{2} + i\left(\frac{\pi}{4} + 2\pi k\right) = \frac{1}{2}\ln 2 + i\left(\frac{\pi}{4} + 2\pi k\right)$
3. Multiply by $(2 + 3i)$: $(2 + 3i)\left[\frac{1}{2}\ln 2 + i\left(\frac{\pi}{4} + 2\pi k\right)\right] = 2 \cdot \frac{1}{2}\ln 2 + 2i\left(\frac{\pi}{4} + 2\pi k\right) + 3i \cdot \frac{1}{2}\ln 2 + 3i^2\left(\frac{\pi}{4} + 2\pi k\right) = \ln 2 - 3\left(\frac{\pi}{4} + 2\pi k\right) + i\left[\frac{3}{2}\ln 2 + 2\left(\frac{\pi}{4} + 2\pi k\right)\right] = \ln 2 - \frac{3\pi}{4} - 6\pi k + i\left(\frac{3}{2}\ln 2 + \frac{\pi}{2} + 4\pi k\right)$.
4. Exponentiate: $(1 + i)^{2 + 3i} = e^{\ln 2 - \frac{3\pi}{4} - 6\pi k} \cdot e^{i(\frac{3}{2}\ln 2 + \frac{\pi}{2} + 4\pi k)} = 2e^{-\frac{3\pi}{4} - 6\pi k}\left[\cos\left(\frac{3\ln 2}{2} + \frac{\pi}{2} + 4\pi k\right) + i\sin\left(\frac{3\ln 2}{2} + \frac{\pi}{2} + 4\pi k\right)\right]$
5. Principal value ($k = 0$): $(1 + i)^{2 + 3i} \approx 2e^{-\frac{3\pi}{4}}\left[\cos\left(\frac{3\ln 2}{2} + \frac{\pi}{2}\right) + i\sin\left(\frac{3\ln 2}{2} + \frac{\pi}{2}\right)\right] \approx 0.2656 \cdot (-0.8539 + 0.5205i) \approx -0.227 + 0.138i$

Properties of Complex Exponentiation

The properties of complex exponentiation reveal subtle nuances due to the multi-valued nature of the complex logarithm and branch cuts.

Property	Formula	Caveat
$z^0$	$1$	For $z \neq 0$
$z^1$	$z$	Trivial identity
$z^{-1}$	$1/z$	The reciprocal of z ($z \neq 0$)
$z^{p+q}$	$z^p \cdot z^q$	It holds only if the branches of $z^p$ and $z^q$ are chosen consistently
$z^{pq}$	$(z^p)^q$	This property fails in general due to branch cuts
$(zw)^p$	$z^p \cdot w^p$	Up to multi-valuedness

1. $z^0 = 1$ (for $z \neq 0$). Any non-zero complex number raised to the power of 0 is defined as 1, consistent with real analysis. z = 0 is excluded because $0^0$ is undefined, e.g., $(3+4i)^0 = 1$
2. $z^1 = z$. Trivial identity —raising a number to the first power returns the number itself, e.g., $(3-4i)^1 = 3-4i$.
3. The reciprocal of z. $z^{-1} = 1/z = \frac{\overline{z}}{|z|^2}$. Implicitly requires $z \ne 0$ (division by zero is undefined), e.g., $(1+i)^{-1} = \frac{1-i}{2}$
4. $z^{p+q} = z^p \cdot z^q$. This exponential law holds only if the branches of $z^p$ and $z^q$ are chosen consistently (i.e., the same integer k). Since $z^w = e^{w\cdot ln(z)}$ and ln(z) is multivalued ($ln(z) = ln|z| + i(\theta + 2k\pi)$), the product $z^p \cdot z^q$ may not equal $z^{p + q}$ if different branches are used.
   z = -1, p = q = 1/2, $z^{p+q}=(-1)^1 = -1$. However, $z^p \cdot z^q = \sqrt{-1} \cdot \sqrt{-1} = (\pm i)(\pm i) = \pm 1$ (multivalued).
5. $z^{pq} = (z^p)^q$. This property fails in general due to branch cuts, e.g., z = -1, p = 2, and q = 1/2, $z^{pq} = (-1)^1 = -1$. $(z^p)^q$, $((-1)^2)^{\frac{1}{2}} = 1^{\frac{1}{2}} = \pm 1$
6. $(zw)^p = z^p \cdot w^p$. This distributive property holds only if the arguments of z and w add without crossing a branch cut. If arg(z) + arg(w) exceeds π or falls below −π, the product zw may lie on a different branch, causing discrepancies.

The Complex Quadratic Formula

Consider the general quadratic in ℂ, $az² + bz + c = 0$ where a, b and c ∈ ℂ, a ≠ 0, the solutions are: $\boxed{z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$ where the square root $\sqrt{b^2 - 4ac}$ is a complex square root!

Derivation by Completing the Square

1. Since a ≠ 0, we can divide by a, $z²+ \frac{b}{a}z + \frac{c}{a} = 0$.
2. Complete the square, $(z+ \frac{b}{2a})² -(\frac{b}{2a})² + \frac{c}{a} = 0$
3. Rearrange to isolate the square: $(z+ \frac{b}{2a})² = \frac{b²-4ac}{4a²}$
4. Take square roots: $z + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$. $w = b^2 - 4ac$ and recall that $w^{\frac{1}{2}} = $ {$\sqrt{w}, -\sqrt{w}$} where $\sqrt{w} = \sqrt{|w|}e^{i\frac{Arg(w)}{2}}$
5. Solve for $z$: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Detailed Example

Solve $z^2 + (2 - 2i)z - (7 + 26i) = 0$:

1. Identify coefficients: $a = 1, b = 2 - 2i, c = -(7 + 26i) = -7 - 26i$
2. Compute the discriminant: $\Delta = b^2 - 4ac = (2 - 2i)^2 - 4(1)(-7 - 26i) = 4 - 8i + 4i^2 + 28 + 104i = 28 + 96i$
3. Find $\sqrt{\Delta} = \sqrt{28 + 96i}$. $|\Delta| = \sqrt{28^2 + 96^2} = \sqrt{784 + 9216} = \sqrt{10000} = 100$. Since the complex number lies in the first quadrant, its argument is given by $\theta = \arctan\left(\frac{96}{28}\right) = \arctan\left(\frac{24}{7}\right) \approx 1.287 \text{ rad}$. The two square root are given by $\sqrt{28 + 96i} = \sqrt{100}(cos(\frac{\theta}{2})+isin(\frac{\theta}{2})) = 10(0.8 + 0.6i) = 8 + 6i$
4. Apply quadratic formula: $z = \frac{-(2 - 2i) \pm (8 + 6i)}{2} = \frac{-2 + 2i \pm (8 + 6i)}{2}$
5. Two solutions: $z_1 = \frac{-2 + 2i + 8 + 6i}{2} = \frac{6 + 8i}{2} = 3 + 4i, z_2 = \frac{-2 + 2i - 8 - 6i}{2} = \frac{-10 - 4i}{2} = -5 - 2i$. Hence, the two roots or solutions of the complex quadratic equation are $z = 3 + 4i$ or $z = -5 - 2i$

The Complex Discriminant

The discriminant $\Delta = b^2 - 4ac$ tells us the geometry of the roots. Unlike real algebra, we must distinguish between whether the coefficients ($a, b, c$) are real or complex.

1. $\Delta > 0$ (Real). Two distinct real numbers, e.g., $z^2 -3z + 2 = 0 \implies (z -1)(z -2) = 0$. Since $\Delta$ is a positive negative number, $\sqrt{\Delta}$ is a positive real number. If a, b, c are real, the expression $\frac{-b\pm \sqrt{\Delta}}{2a}$ is just adding/subtracting real numbers, resulting in two distinct real numbers.
2. $\Delta = 0$ (The Degenerate Case). One repeated real root (x = -b/2a). Even with complex coefficients, if the discriminant is zero, the two roots “collapse” into one point with multiplicity 2, e.g., $(z - (1+i))^2 = z^2 -2(1+i)z + 2i$. Here Δ = 0, and the single root is 1 + i (repeated).
3. $\Delta < 0$ (Real). When coefficients $a, b, c$ are real and $\Delta$ is negative, the square root $\sqrt{\Delta}$ becomes purely imaginary: $i\sqrt{|\Delta|}$. The roots are $z = \frac{-b}{2a} \pm i\frac{\sqrt{|\Delta|}}{2a}$. Because the real parts are identical and imaginary parts are opposite, the roots are complex conjugates.
4. $\Delta \in \mathbb{C} \setminus \mathbb{R}$ (Non-Real Discriminant). If any coefficient is complex, $\Delta$ is often complex. This breaks the symmetry required for the Conjugate Root Theorem. The , e.g. $z^2 + (1+i) = 0, Δ = -4 -4i$ (this is complex, not real).​ Roots: $z \approx \pm(0.455 - 1.099i)$. Root 2 is not the conjugate of Root 1.

One root is the negative of the other, but they are not conjugates of each other. The Conjugate Root Theorem only guarantees conjugate pairs if the original polynomial has strictly real coefficients.