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Recall: Relationship Between Complex Differentiability and the Cauchy-Riemann Equations

Necessary Condition: Complex Differentiability Implies Cauchy-Riemann Equations

If a function f(z) = u(x, y) + iv(x, y) is complex-differentiable at a point z₀ =x₀ +iy₀, then its real and imaginary parts, u(x,y) and v(x,y), must satisfy the Cauchy-Riemann equations at that point: $\frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0), \frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0)$.

Sufficient Condition: Cauchy-Riemann Equations (with Continuous Partial Derivatives) Imply Complex Differentiability

Proposition. Let f(z) = u(x, y) + iv(x, y) for z = x + iy ∈ D where D ⊆ ℂ is an open set. Assume that u and v have continuous first partial derivatives throughout D and they satisfy the Cauchy-Riemann equations at a point z ∈ D. Then, f'(z) exists, i.e., f is differentiable at z.

Analytic functions

Definition. A complex function f is said to be analytic (holomorphic) at a point z₀ if it satisfies any of the following equivalent conditions:

Differentiable in a neighborhood. There exists an open neighborhood U of z₀ such that f is complex differentiable at every point in U. Complex differentiability means that the derivative f′(z₀) exists for all z₀ ∈ U in the complex sense, f′(z₀) = $\lim_{h \to 0} \frac{f(z₀+h)-f(z₀)}{h}$.
Here the limit must be the same regardless of the direction from which h approaches zero in the complex plane.
Open disc criterion. There exists some radius r > 0 such that f is complex differentiable at every point in the open disc B(z₀ ; r) = {z: |z - z₀| < r}. This is a specific case of Condition 1, as an open disc is a specific type of open neighborhood around z₀.
Power series expansion. The function can be locally expressed as a convergent power series:$f(z) = \sum_{n = 0}^\infty a_n(z -z₀)^n$ in some neighborhood of z₀. This series converges absolutely and uniformly on compact subsets within its disk of convergence.

The Complex Exponential Function

The complex exponential function, denoted as eᶻ or exp(z), is one of the most important functions in all of mathematics. It is the unique entire function that extends the real exponential function eˣ to the complex plane. This function is not only entire but also periodic.

To define the complex exponential function, we seek a function f: ℂ→ℂ that satisfies two key properties:

f(z₁ + z₂) = f(z₁)·f(z₂) for all z₁, z₂ ∈ ℂ
Consistency with the Real Exponential: f(x) = eˣ where x is a real number.

Let z = x + iy, where x, y ∈ ℝ. f(z) = f(x + iy) =[1] f(x)·f(iy) =[2] eˣ·f(iy)

Writing f(iy) = A(y) + iB(y), f(z) = eˣ·A(y) + ieˣ·B(y), where A(y) and B(y) are real-valued functions of y.

For f(z) to be a well-behaved function, we want it to be differentiable everywhere, which means it must satisfy the Cauchy-Riemann equations. The real and imaginary parts of f(z) are: u(x, y) = eˣ·A(y), v(x, y) = eˣ·B(y). The partial derivatives are uₓ = eˣ·A(y), u_y = eˣ·A’(y), vₓ = eˣ·B(y), and v_y = eˣ·B’(y).

Applying the Cauchy-Riemann equations, uₓ = v_y and u_y = - vₓ, we get: eˣ·A(y) = eˣ·B’(y), then, A(y) = B’(y). Similarly, eˣ·A’(y) = - eˣ·B(y), then A’(y) = -B(y).

From these two equations, we can derive a second-order differential equation for B(y): B’’(y) = -B(y) ↭ B’’(y) + B(y) = 0.

This linear, homogeneous, constant-coefficient ODE has characteristic equation r² + 1 = 0 ⇒ r = ±i. Complex Conjugate Roots: If the roots are complex conjugates a ± bi, the general solution is $y(t) = e^{at}(αcos(bt) + βsin(bt))$, where ‘α’ and ‘β’ are real numbers.

The general solution to our equation (a = 0, b = 1) is B(y) = αcos(y) + βsin(y) for some constants α, β ∈ ℝ. Using A(y) = B’(y), we find A(y) = -αsin(y) + βcos(y).

Now, we use the initial condition from the real exponential. Since f(0) = e⁰ = 1 =[f(z) = eˣ·A(y) + ieˣ·B(y)] e⁰A(0) + ie⁰B(0) ⇒ A(0) = 1, B(0) = 0.

From A(0) = 1 ⇒[-αsin(y) + βcos(y)] -αsin(0) + βcos(0) = 1 ⇒ β = 1.

From B(0) = 0 ⇒[B(y) = αcos(y) + βsin(y)] αcos(0) + βsin(0) = 0 ⇒ α = 0. Thus, we have uniquely determined the functions A(y) and B(y): B(y) = sin(y), A(y) = cos(y). Substituting these back into the expression for f(z), we arrive at the definition of the complex exponential function, f(z) = eˣ·cos(y) + ieˣ·sin(y) = eˣ(cos(y) + isin(y)).

Properties

The function is well-defined for all z ∈ ℂ.
As shown through the derivation using the Cauchy-Riemann equations, it is an entire function, meaning it is analytic everywhere in the complex plane:

You may check the Cauchy–Riemann equations directly from the definition f(z) = eˣ(cos(y) + isin(y)) and find that the partial derivatives of the real and imaginary parts are continuous and satisfy the Cauchy–Riemann equations at every point, so the function is holomorphic on the whole plane.
exp(z) is entired because it can be expressed as $e^z = \sum_{n=0}^∞ \frac{z^n}{n!}$. This is a single power series centered at z₀ = 0 with coefficients $a_n = \frac{1}{n!}$.

To determine where this series converges, we use the ratio test. Let the power series be $\sum_{n=0}^∞ aₙ(x-c)ⁿ$. The ratio test involves computing: $L = \lim_{n \to ∞} |\frac{aₙ₊₁}{aₙ}|$. If L < 1, the series converges.

$L = \lim_{n \to ∞} |\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}| = \lim_{n \to ∞} |\frac{n!}{(n+1)!}| = \lim_{n \to ∞} \frac{1}{n+1} = 0 $ < 1, the series converges absolutely for any fixed z ∈ ℂ, the radius of convergence is R = 1/L = 1/0 = ∞. Because R = ∞, the series converges for all $z \in \mathbb{C}$ making eᶻ entire (analytic everywhere in the complex plane).

Addition Formula: $e^{z_1+z_2} = e^{z_1}⋅e^{z_2}$

$e^{z_1}⋅e^{z_2} =[\text{By definition}] e^{x_1}(cos(y_1) + isin(y_1))⋅e^{x_2}(cos(y_2) + isin(y_2)) = e^{(x_1+x_2)}[(cos(y_1)cos(y_2)-sin(y_1)sin(y_2) + i(sin(y_1)cos(y_2)+cos(y_1)sin(y_2)))] =$

Using the angle addition formulas for cosine and sine: $ = e^{x_1+x_2}(cos(y_1+y_2)+isin(y_1+y_2)) = e^{(x_1+x_2)+i(y_1+y_2)} = e^{z_1+z_2}$

Consistency with the Real Exponential. If z is real, then exp(z) = exp(x + i0) =[By definition] eˣ(cos(0) + isin(0)) = eˣ
Modulus and Argument: |eᶻ| = $|e^x(cos(y)+isin(y)) = e^x\sqrt{cos^2(y)+sin^2(y)}=$eˣ. The argument of eᶻ is any angle whose cosine and sine are cos(y) and sin(y) respectively. Because cosine and sine are periodic with period 2π, every number of the form y + 2kπ, k ∈ ℤ is an admissible argument. Thus, arg(z) = y +2kπ for any integer k. This shows that the real part of z controls the magnitude of eᶻ, while the imaginary part controls its angle.
Non-Zero Property. Since |eᶻ| = eˣ and eˣ > 0 for all real x, it follows that eᶻ ≠ 0 for all z ∈ ℂ.
Periodicity: The complex exponential function is periodic with period 2πi. $e^{z+2k\pi·i} = e^{z}e^{2k\pi·i} = e^{z}(cos(2k\pi·i)+isin(2k\pi·i)) = e^{z}(1 + i·0) = e^{z}$
Solutions to eᶻ = α. For any non-zero complex number α, the equation eᶻ = α has infinitely many solutions.

If we write α in polar form as $α = e^{i\theta}$ where R = |α| and θ = arg(α), then the solutions for eᶻ = α are given by: eˣ = R ⇒ x = ln(R) and y = θ + 2πk, k ∈ ℤ. Therefore, the solutions are ln(|α|) + i(arg(α) + 2πk), for any integer k ∈ ℤ.

Example: eᶻ = 2 + 2i, 2 + 2i = $2\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}) = 2\sqrt{2}(cos(\frac{\pi}{4}+isin(\frac{\pi}{4}))) = 2\sqrt{2}(cos(2n\pi+\frac{\pi}{4})+isin(2n\pi+\frac{\pi}{4}))$ where n ∈ ℤ

take eˣ $=2\sqrt{2}, \text{i.e.,} x = ln(2\sqrt{2}), y = 2n\pi+\frac{\pi}{4}$

Then, eᶻ = $e^{ln(2\sqrt{2})+i(2n\pi+\frac{\pi}{4})}$ = 2 + 2i where n ∈ ℤ. Therefore, the equation eᶻ = 2 + 2i has infinitely many solutions.

The derivative of the exponential function is quite simple: $\frac{d}{dz}e^z = e^z$.

If f is an analytic function with real and imaginary parts f(z) = eᶻ = u(x, y) + iv(x, y) = eˣcos(y) + ieˣsin(y), then $f'(z) = (eᶻ)' = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = eˣcos(y) + ieˣsin(y) = eᶻ$

An alternative proof is as follows.

Theorem. If a function f(z) can be represented by a power series, f(z) = $\sum_{n=0}^∞a_n(z-z_0)^n$, that converges in a disk ∣z − z₀∣ < R (with R > 0), then f is analytic within that disk, and its derivative is found by differentiating the power series term by term: f’(z) = $\sum_{n=1}^∞na_n(z-z_0)^{n-1}$ (we are using the power rule for differentiation on each term, $\frac{d}{dz}a_n(z-z_0)^n = na_n(z-z_0)^{n-1}$ and the term for n = 0 is a constant, a₀, so its derivative is zero).

In our case, $e^z = \sum_{n=0}^∞ \frac{z^n}{n!}$ center at z₀ = 0 with $a_n = \frac{1}{n!}$ and R = = ∞. Thus, term-by-term differentiation is valid for all z ∈ ℂ.

Differentiate the series term by term: $\frac{d}{dz}e^z = \frac{d}{dz}(\sum_{n=0}^∞ \frac{z^n}{n!}) = \sum_{n=0}^∞ \frac{d}{dz}(\frac{z^n}{n!})$

For n = 0, $\frac{d}{dz}(\frac{z^0}{0!}) = \frac{d}{dz}(1) = 0$. For n ≥ 1, $\frac{d}{dz}(\frac{z^n}{n!}) = \frac{nz^{n-1}}{n!} = \frac{z^{n-1}}{(n-1)!}$. The series becomes: $\sum_{n=0}^∞ \frac{d}{dz}(\frac{z^n}{n!}) = \sum_{n=1}^∞ \frac{z^{n-1}}{(n-1)!} = [\text{Reindexing the Series, k = n-1}] \sum_{k=0}^∞ \frac{z^{k}}{k!} = e^z$ ∎

The derivative of eᶻ is identical to itself. Since eᶻ is entire, repeated differentiation yields: $\frac{d^2}{dz^2}eᶻ = eᶻ, \frac{d^3}{dz^3}eᶻ = eᶻ, \dots$ confirming eᶻ is “smooth” (C^∞) and infinitely differentiable in ℂ, meaning having derivatives of all orders that behave nicely.

eᶻ solves the first-order differential equation $\frac{df}{dz} = f(z)$ with initial condition f(0) = 1.

The given differential equation is a classic example of a separable differential equation. The equation can be rewritten as $\frac{df}{f} = dz$. Integrating both sides gives: $\int \frac{df}{f} = \int dz \leadsto ln|f| = z + C$

To solve for f, we exponentiate both sides: $|f| = e^{z + C} = e^{z}e^{C}$. Let’s rename the constant $e^{C}$ as A. Since $e^{C}$ is always positive, A can be any non-zero real number. The general solution is: f(z) = Aeᶻ.

Now, we use the initial condition f(0)=1 to find the value of A: f(0) = Ae⁰ = A·1 = A, hence A = 1. Plugging this value back into the general solution gives us the specific solution: f(z) = Aeᶻ = 1eᶻ = eᶻ. This is the unique solution to the initial value problem.

Euler’s formula eⁱʸ = cos(y) + isin(y). A special case of the complex exponential function, when the real part of the exponent is zero, leads to one of the most celebrated formulas in mathematics, Euler’s Formula. Setting x=0 in the definition of eᶻ, we get: eⁱʸ = e⁰(cos(y) + isin(y)) = cos(y) + isin(y).

This formula establishes a profound connection between the exponential function and the trigonometric functions. It shows that the complex exponential function can be viewed as a generalization of the unit circle in the complex plane. As the real variable y increases, the point eⁱʸ traces out the unit circle ∣z∣=1 in a counterclockwise direction, starting from eⁱ⁰ = 1.

This provides a powerful and compact way to represent complex numbers in polar form. Any complex number z = x + iy can be written as z = r(cosθ + isinθ), where r = ∣z∣ and θ = arg(z). Using Euler’s formula, this become $z = re^{i\theta}$. This exponential form is extremely useful for performing operations like multiplication, division, and finding powers and roots of complex numbers, e.g., $z_1·z_2 = (r_1e^{i\theta_1})(r_2e^{i\theta_2}) = r_1r_2e^{i(\theta_1+\theta_1)}$ (to multiply two complex numbers, we multiply their moduli and add their arguments.)

Complex Trigonometric Functions

Euler’s formula leads to elegant expressions for the trigonometric functions. Euler’s formula states that $e^{i y} = \cos(y) + i·\sin(y)$. If we also write its complex conjugate $e^{-i y} = \cos(y) - i·\sin(y),$ then by adding and subtracting these two expressions we isolate cosine and sine. $cos(y) = \frac{eⁱʸ+e⁻ⁱʸ}{2}, sin(y) = \frac{eⁱʸ-e⁻ⁱʸ}{2i}$.
These definitions are consistent with the real-valued functions when z is a real number. Since the exponential function eᶻ is an entire function, and sums, products, and compositions of entire functions are also entire, it follows that cos(z) and sin(z) are entire functions. They are analytic everywhere in the complex plane.
Derivatives of Sine and Cosine. The derivatives of the complex sine and cosine functions can be found by differentiating their exponential definitions. Using the chain rule and the fact that $\frac{d}{dz}e^z = e^z$, we have $\frac{d}{dz}cos(z) = \frac{d}{dz}\frac{eⁱᶻ+e⁻ⁱᶻ}{2} = \frac{ieⁱᶻ-ie⁻ⁱᶻ}{2} = i\frac{eⁱᶻ-e⁻ⁱᶻ}{2} = -\frac{eⁱᶻ-e⁻ⁱᶻ}{2i} = -sin(z)$. Similarly, for the sine function: $\frac{d}{dz}sin(z) = \frac{d}{dz}\frac{eⁱᶻ-e⁻ⁱᶻ}{2i} = \frac{ieⁱᶻ+ie⁻ⁱᶻ}{2i} = \frac{eⁱᶻ+e⁻ⁱᶻ}{2} = cos(z)$. These results are identical to the familiar formulas from real calculus and extend naturally to the complex domain.
Trigonometric Identities in the Complex Plane. All the standard trigonometric identities that hold for real arguments also hold for complex arguments. For example, the Pythagorean identity, $cos^2(z) + sin^2(z) = (\frac{eⁱᶻ+e⁻ⁱᶻ}{2})^2 + (\frac{eⁱᶻ-e⁻ⁱᶻ}{2i})^2 = \frac{e²ⁱᶻ+ 2 + e⁻²ⁱᶻ}{4} + \frac{e²ⁱᶻ -2 +e⁻²ⁱᶻ}{-4} = \frac{e²ⁱᶻ+ 2 + e⁻²ⁱᶻ -e²ⁱᶻ + 2 - e⁻²ⁱ}{4} = \frac{4}{4} = 1$. Similarly, the angle addition formulas hold: $\sin(z_1 + z_2) = \sin z_1 \cos z_2 + \cos z_1 \sin z_2, \cos(z_1 + z_2) = \cos z_1 \cos z_2 - \sin z_1 \sin z_2, \sin(z_1 - z_2) = \sin z_1 \cos z_2 - \cos z_1\sin z_2$

Mapping Properties of the Complex Exponential

Real axis (y = 0). $\\{x+0i : x \in ℝ\\} \xrightarrow{\exp} \\{e^x : x \in ℝ\\} = (0,\infty).$ All positive real numbers.
Imaginary axis (x = 0). $\\{0+iy : y \in ℝ\\} \xrightarrow{\exp} \\{e^{i y} : y \in ℝ\\} = S^1$.The unit circle.
Horizontal line (y = $y_0$), $\\{x+i y_0 : x \in ℝ \\} \xrightarrow{\exp} \\{e^x e^{i y_0} : x \in ℝ \\}$. A half‐line (ray) without the origin at argument $y_0$, with radius running from 0 to + $\infty$.
Vertical line ( x = $x_0$), $\\{x_0+i y : y \in ℝ \\}\xrightarrow{\exp}\\{e^{x_0}e^{iy} : y\in\R\\}$. A full circle of radius $e^{x_0}$ centered at the origin.
Strip $0 \le y < 2\pi$, the image is $ℂ \setminus \\{0\\}$, the entire punctured plane.
Strip $\alpha < y < \beta$, the image is the sector between two rays.
Left half‐plane x < 0, the image is $\\{w:0<|w|<1\\}$, the open unit disk minus zero.
Right half‐plane x > 0, the image is $\\{w:|w|>1\\}$, the exterior of the unit disk.