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∑

∂

∞

ℂ

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Course 3

Elementary Analytic Functions

Study the exponential, logarithmic, trigonometric, hyperbolic, and power functions in the complex domain with their properties, branch cuts, and applications.

Intermediate5-6 hours10 Practice Questions

Learning Objectives

Understand the complex exponential function and its properties

Master the complex logarithm and its branches

Study complex trigonometric and inverse trigonometric functions

Learn about complex hyperbolic functions

Understand power functions and their multi-valued nature

Work with branch cuts and principal values

Apply properties of elementary functions to solve problems

Understand the relationship between real and complex elementary functions

The Complex Exponential Function

Definition 3.1: Complex Exponential

The complex exponential function is defined as:

e^z = e^{x + iy} = e^x(\cos y + i\sin y) = e^x e^{iy}

for

z = x + iy \in \mathbb{C}

Theorem 3.1: Properties of Complex Exponential

The complex exponential satisfies:

$e^{z_1 + z_2} = e^{z_1} e^{z_2}$ (addition formula)
$e^z \neq 0$ for all $z$
$e^{z + 2\pi i} = e^z$ (periodic with period $2\pi i$ )
$(e^z)' = e^z$ (its own derivative)
$|e^z| = e^x$ and $\arg(e^z) = y + 2k\pi$

Proof of Theorem 3.1:

Using the definition: $e^{z_1 + z_2} = e^{(x_1 + x_2) + i(y_1 + y_2)} = e^{x_1 + x_2}[\cos(y_1 + y_2) + i\sin(y_1 + y_2)]$

Using trigonometric identities: $= e^{x_1}e^{x_2}[(\cos y_1\cos y_2 - \sin y_1\sin y_2) + i(\sin y_1\cos y_2 + \cos y_1\sin y_2)]$

$= e^{x_1}(\cos y_1 + i\sin y_1) \cdot e^{x_2}(\cos y_2 + i\sin y_2) = e^{z_1} e^{z_2}$ ✓

Periodicity: $e^{z + 2\pi i} = e^z e^{2\pi i} = e^z(\cos 2\pi + i\sin 2\pi) = e^z$ ✓

∎

Example 3.1

Problem: Compute $e^{1 + i\pi}$ .

Solution:

$e^{1 + i\pi} = e^1(\cos\pi + i\sin\pi) = e(-1 + 0i) = -e$

The Complex Logarithm

Definition 3.2: Complex Logarithm

w

satisfies

e^w = z

for

z \neq 0

, then

w

is called a logarithm of

z

, denoted

\text{Ln } z

. Writing

z = re^{i\theta}

and

w = u + iv

\text{Ln } z = \ln r + i(\theta + 2k\pi) = \ln|z| + i\arg z

where

k \in \mathbb{Z}

. The principal value is

\ln z = \ln|z| + i\arg z

with

\arg z \in (-\pi, \pi]

Remark 3.1

The complex logarithm is infinitely many-valued because the argument can differ by multiples of

2\pi

. The principal branch corresponds to

k = 0

and a branch cut along the negative real axis.

Example 3.2

Problem: Find all values of $\text{Ln}(-1)$ .

Solution:

Since $-1 = e^{i\pi}$ , we have $\text{Ln}(-1) = \ln 1 + i(\pi + 2k\pi) = i\pi(2k + 1)$ for $k \in \mathbb{Z}$ .

The principal value is $\ln(-1) = i\pi$ (for $k = 0$ ).

Trigonometric Functions

Definition 3.3: Complex Trigonometric Functions

For

z \in \mathbb{C}

\sin z = \frac{e^{iz} - e^{-iz}}{2i}, \quad \cos z = \frac{e^{iz} + e^{-iz}}{2}

Other trigonometric functions are defined as:

\tan z = \frac{\sin z}{\cos z}

\cot z = \frac{\cos z}{\sin z}

, etc.

Theorem 3.2: Properties of Complex Trigonometric Functions

Complex trigonometric functions satisfy:

$\sin(z + 2\pi) = \sin z$ , $\cos(z + 2\pi) = \cos z$ (period $2\pi$ )
$\cos^2 z + \sin^2 z = 1$
$\sin z$ and $\cos z$ are unbounded (unlike real versions)
$(\sin z)' = \cos z$ , $(\cos z)' = -\sin z$

Example 3.3

Problem: Show that $|\sin(iy)|$ is unbounded.

Solution:

$\sin(iy) = \frac{e^{i(iy)} - e^{-i(iy)}}{2i} = \frac{e^{-y} - e^{y}}{2i} = i\sinh y$

Therefore $|\sin(iy)| = |\sinh y| = \frac{e^{|y|} - e^{-|y|}}{2} \to \infty$ as $|y| \to \infty$ .

Hyperbolic Functions

Definition 3.4: Complex Hyperbolic Functions

For

z \in \mathbb{C}

\sinh z = \frac{e^z - e^{-z}}{2}, \quad \cosh z = \frac{e^z + e^{-z}}{2}

Theorem 3.3: Relationship to Trigonometric Functions

Hyperbolic and trigonometric functions are related by:

\sinh(iz) = i\sin z, \quad \cosh(iz) = \cos z, \quad \sin(iz) = i\sinh z, \quad \cos(iz) = \cosh z

Power Functions

Definition 3.5: Complex Power Function

For

z \neq 0

and

\alpha \in \mathbb{C}

z^\alpha = e^{\alpha \text{Ln } z} = e^{\alpha(\ln|z| + i\arg z + 2k\pi i)}

The principal value is

z^\alpha = e^{\alpha \ln z}

where

\ln z

is the principal logarithm.

Example 3.4

Problem: Find all values of $i^i$ .

Solution:

$i^i = e^{i\text{Ln } i} = e^{i(\ln 1 + i(\pi/2 + 2k\pi))} = e^{i^2(\pi/2 + 2k\pi)} = e^{-\pi/2 - 2k\pi}$

So $i^i$ takes infinitely many real values: $e^{-\pi/2}, e^{-5\pi/2}, e^{-9\pi/2}, \ldots$

Inverse Trigonometric Functions

Definition 3.6: Inverse Trigonometric Functions

The inverse trigonometric functions are defined as:

\arcsin z = -i\text{Ln}(iz + \sqrt{1-z^2}), \quad \arccos z = -i\text{Ln}(z + \sqrt{z^2-1})

\arctan z = \frac{i}{2}\text{Ln}\left(\frac{1-iz}{1+iz}\right)

These are multi-valued functions. Their principal branches are obtained by choosing appropriate branches of the square root and logarithm.

Example 3.5

Problem: Find all values of $\arcsin 2$ .

Solution:

$\arcsin 2 = -i\text{Ln}(2i + \sqrt{1-4}) = -i\text{Ln}(2i + i\sqrt{3}) = -i\text{Ln}(i(2+\sqrt{3}))$

$= -i[\ln(2+\sqrt{3}) + i(\pi/2 + 2k\pi)] = \pi/2 + 2k\pi + i\ln(2+\sqrt{3})$

Note that $\arcsin 2$ is not a real number, unlike in real analysis.

Theorem 3.4: Properties of Inverse Functions

$\sin(\arcsin z) = z$ (on appropriate branches)
$\cos(\arccos z) = z$
$\tan(\arctan z) = z$
The derivatives are: $(\arcsin z)' = \frac{1}{\sqrt{1-z^2}}$ , $(\arccos z)' = -\frac{1}{\sqrt{1-z^2}}$ , $(\arctan z)' = \frac{1}{1+z^2}$

Hyperbolic Functions

Definition 3.7: Hyperbolic Functions

The hyperbolic functions are defined as:

\sinh z = \frac{e^z - e^{-z}}{2}, \quad \cosh z = \frac{e^z + e^{-z}}{2}

\tanh z = \frac{\sinh z}{\cosh z}, \quad \coth z = \frac{\cosh z}{\sinh z}

Theorem 3.5: Relationships with Trigonometric Functions

\sinh(iz) = i\sin z, \quad \cosh(iz) = \cos z

\sin(iz) = i\sinh z, \quad \cos(iz) = \cosh z

Example 3.6

Problem: Show that $\cosh^2 z - \sinh^2 z = 1$ .

Solution:

\cosh^2 z - \sinh^2 z = \left(\frac{e^z + e^{-z}}{2}\right)^2 - \left(\frac{e^z - e^{-z}}{2}\right)^2 = \frac{e^{2z} + 2 + e^{-2z} - (e^{2z} - 2 + e^{-2z})}{4} = 1

Example 3.7

Problem: Find all zeros of $\sinh z$ .

Solution:

$\sinh z = 0$ implies $e^z - e^{-z} = 0$ , so $e^{2z} = 1$

This gives $2z = 2k\pi i$ , so $z = k\pi i$ for $k \in \mathbb{Z}$

Therefore, $\sinh z = 0$ if and only if $z = k\pi i$ .

Multi-valued Functions and Branch Cuts

Definition 3.8: Branch

A branch of a multi-valued function is a single-valued function obtained by restricting the function to a domain where it is single-valued and continuous. A branch cut is a curve (or line) along which a branch is discontinuous.

Example 3.8

Problem: Describe the principal branch of $\text{Ln } z$ .

Solution:

The principal branch of $\text{Ln } z$ is $\ln z = \ln|z| + i\arg z$ where $-\pi < \arg z \leq \pi$ .

This branch is analytic in the cut plane $\mathbb{C} \setminus (-\infty, 0]$ (the negative real axis is the branch cut).

Across the branch cut, the function jumps by $2\pi i$ .

Example 3.9

Problem: Find the branch points of $f(z) = \sqrt{z(z-1)}$ .

Solution:

A branch point is where the function becomes multi-valued regardless of how small a neighborhood we take.

At $z = 0$ and $z = 1$ , encircling these points changes the value of $\sqrt{z(z-1)}$ .

Also, $z = \infty$ is typically a branch point for square roots.

A common branch cut is the line segment from $0$ to $1$ .

Remark 3.1

The concept of Riemann surfaces provides a geometric way to understand multi-valued functions. On a Riemann surface, each branch of a multi-valued function corresponds to a "sheet", and the sheets are connected along branch cuts.

Additional Worked Examples

Example 3.10

Problem: Evaluate $\ln(-1)$ (all values).

Solution:

$\ln(-1) = \ln|−1| + i\arg(-1) = \ln 1 + i(\pi + 2k\pi) = i\pi(1 + 2k)$

Principal value: $\ln(-1) = i\pi$

All values: $\ldots, -3i\pi, -i\pi, i\pi, 3i\pi, 5i\pi, \ldots$

Example 3.11

Problem: Show that $e^{z_1} \cdot e^{z_2} = e^{z_1 + z_2}$ .

Solution:

Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ . Then:

e^{z_1} \cdot e^{z_2} = e^{x_1 + iy_1} \cdot e^{x_2 + iy_2} = e^{x_1}e^{iy_1} \cdot e^{x_2}e^{iy_2} = e^{x_1 + x_2}e^{i(y_1 + y_2)} = e^{z_1 + z_2}

Example 3.12

Problem: Find all cube roots of $-8i$ .

Solution:

Write $-8i = 8e^{-i\pi/2} = 8e^{i(-\pi/2 + 2k\pi)}$

Cube roots: $w_k = 8^{1/3} e^{i(-\pi/2 + 2k\pi)/3} = 2e^{i(-\pi/6 + 2k\pi/3)}$ for $k = 0, 1, 2$

$w_0 = 2e^{-i\pi/6} = \sqrt{3} - i$ , $w_1 = 2e^{i\pi/2} = 2i$ , $w_2 = 2e^{i7\pi/6} = -\sqrt{3} - i$

Study Tips

1. Handling Multi-valued Functions

Always specify which branch you're using. The principal branch uses $-\pi < \arg z \leq \pi$ for logarithms and square roots. Be careful with branch cuts when integrating.

2. Exponential Identities

Many identities from real analysis extend: $e^{z_1 + z_2} = e^{z_1}e^{z_2}$ , $(e^z)' = e^z$ , $e^{z + 2k\pi i} = e^z$ . However, $(e^z)^w \neq e^{zw}$ in general.

3. Trigonometric Functions

Remember that $|\sin z|$ and $|\cos z|$ can exceed 1 for complex arguments. Use Euler's formula to derive identities.

Additional Examples

Example 3.13

Problem: Find all values of $\ln(1+i)$ .

Solution:

$|1+i| = \sqrt{2}$ , $\arg(1+i) = \pi/4$

$\ln(1+i) = \ln\sqrt{2} + i(\pi/4 + 2k\pi) = \frac{1}{2}\ln 2 + i(\pi/4 + 2k\pi)$

Principal value: $\frac{1}{2}\ln 2 + i\pi/4$

Example 3.14

Problem: Evaluate $\sqrt{i}$ (all values).

Solution:

$\sqrt{i} = e^{(1/2)\text{Ln } i} = e^{(1/2)(\ln 1 + i(\pi/2 + 2k\pi))} = e^{i(\pi/4 + k\pi)}$

For $k = 0$ : $e^{i\pi/4} = \frac{1+i}{\sqrt{2}}$

For $k = 1$ : $e^{i5\pi/4} = -\frac{1+i}{\sqrt{2}}$

So $\sqrt{i} = \pm\frac{1+i}{\sqrt{2}}$

Example 3.15

Problem: Find the derivative of $f(z) = z^i$ (principal branch).

Solution:

Using $z^i = e^{i\ln z}$ (principal branch):

$f'(z) = e^{i\ln z} \cdot i \cdot \frac{1}{z} = iz^i \cdot \frac{1}{z} = iz^{i-1}$

Note: This formula holds for the principal branch in the cut plane.

Example 3.16

Problem: Show that $\cos(z_1 + z_2) = \cos z_1 \cos z_2 - \sin z_1 \sin z_2$ .

Solution:

Using Euler's formula:

$\cos(z_1 + z_2) = \frac{e^{i(z_1+z_2)} + e^{-i(z_1+z_2)}}{2} = \frac{e^{iz_1}e^{iz_2} + e^{-iz_1}e^{-iz_2}}{2}$

Expanding and using definitions of cosine and sine:

$= \cos z_1 \cos z_2 - \sin z_1 \sin z_2$

Example 3.17

Problem: Find all solutions to $\sin z = 2$ .

Solution:

Write $\sin z = \frac{e^{iz} - e^{-iz}}{2i} = 2$

Let $w = e^{iz}$ , then $\frac{w - 1/w}{2i} = 2$ , so $w - 1/w = 4i$

Multiplying: $w^2 - 4iw - 1 = 0$

$w = 2i \pm \sqrt{-4 + 1} = 2i \pm i\sqrt{3} = i(2 \pm \sqrt{3})$

So $e^{iz} = i(2 \pm \sqrt{3})$ , giving $iz = \ln(i(2 \pm \sqrt{3})) = \ln(2 \pm \sqrt{3}) + i(\pi/2 + 2k\pi)$

Therefore: $z = \frac{\pi}{2} + 2k\pi - i\ln(2 \pm \sqrt{3})$

Example 3.18

Problem: Prove that $|e^z| = e^{\text{Re}(z)}$ .

Solution:

If $z = x + iy$ , then $e^z = e^x e^{iy}$

$|e^z| = |e^x| \cdot |e^{iy}| = e^x \cdot 1 = e^{\text{Re}(z)}$

This shows that $|e^z|$ depends only on the real part of $z$ .

Example 3.19

Problem: Find the period of $\tan z$ .

Solution:

Since $\sin z$ and $\cos z$ have period $2\pi$ , $\tan z = \sin z/\cos z$ also has period $2\pi$ (or more precisely, period $\pi$ since $\tan(z + \pi) = \tan z$ ).

Example 3.20

Problem: Show that $\cosh^2 z - \sinh^2 z = 1$ .

Solution:

\cosh^2 z - \sinh^2 z = \left(\frac{e^z + e^{-z}}{2}\right)^2 - \left(\frac{e^z - e^{-z}}{2}\right)^2 = \frac{e^{2z} + 2 + e^{-2z} - (e^{2z} - 2 + e^{-2z})}{4} = 1

Example 3.21

Problem: Find all solutions to $e^z = -1$ .

Solution:

$e^z = -1 = e^{i\pi}$

$z = i\pi + 2k\pi i = i\pi(1 + 2k)$ for $k \in \mathbb{Z}$

So $z = \ldots, -3i\pi, -i\pi, i\pi, 3i\pi, \ldots$

Example 3.22

Problem: Show that $\sin z = 0$ if and only if $z = k\pi$ for some integer $k$ .

Solution:

$\sin z = 0$ means $\frac{e^{iz} - e^{-iz}}{2i} = 0$ , so $e^{iz} = e^{-iz}$

This implies $e^{2iz} = 1 = e^{2k\pi i}$ , so $2iz = 2k\pi i$ , giving $z = k\pi$ .

Example 3.23

Problem: Show that $\cos z = 0$ if and only if $z = \pi/2 + k\pi$ for some integer $k$ .

Solution:

$\cos z = 0$ means $\frac{e^{iz} + e^{-iz}}{2} = 0$ , so $e^{iz} = -e^{-iz}$

This implies $e^{2iz} = -1 = e^{i\pi}$ , so $2iz = i\pi + 2k\pi i$

Therefore: $z = \pi/2 + k\pi$ .

Example 3.24

Problem: Find the derivative of $f(z) = z^{1/3}$ (principal branch).

Solution:

Using $z^{1/3} = e^{(1/3)\ln z}$ :

$f'(z) = e^{(1/3)\ln z} \cdot \frac{1}{3} \cdot \frac{1}{z} = \frac{1}{3}z^{1/3} \cdot \frac{1}{z} = \frac{1}{3}z^{-2/3}$

This is valid in the cut plane where the principal branch of the logarithm is defined.

Example 3.25

Problem: Determine the branch cut for $f(z) = \sqrt{z^2 - 1}$ and explain how to choose it to make the function analytic.

Solution:

The branch points occur where $z^2 - 1 = 0$ , i.e., at $z = \pm 1$ .

We can write $f(z) = \sqrt{(z-1)(z+1)}$ . To make this single-valued, we need a branch cut connecting the branch points.

Option 1: Cut along the real axis from $-1$ to $1$ . This makes the function analytic in $\mathbb{C} \setminus [-1, 1]$ .

Option 2: Cut along rays from $-1$ to $-\infty$ and from $1$ to $\infty$ along the real axis. This makes the function analytic in $\mathbb{C} \setminus ((-\infty, -1] \cup [1, \infty))$ .

The choice depends on the application. For many purposes, cutting along $[-1, 1]$ is convenient as it keeps the function analytic on most of the real axis.

Example 3.26

Problem: Show how different branch cuts for $\ln z$ affect the value of $\ln(-1)$ .

Solution:

The general form is $\ln(-1) = \ln|{-1}| + i\arg(-1) = 0 + i(\pi + 2k\pi) = i\pi(2k+1)$ for any integer $k$ .

Principal branch (cut along negative real axis): $\arg(-1) = \pi$ , so $\ln(-1) = i\pi$

Alternative branch (cut along positive real axis): If we choose $\arg z \in (0, 2\pi]$ , then $\arg(-1) = \pi$ still, so $\ln(-1) = i\pi$

Different branch: If we choose $\arg z \in (\pi/2, 5\pi/2]$ , then $\arg(-1) = 3\pi$ , so $\ln(-1) = 3i\pi$

The branch cut determines which values of the argument are used, thus affecting the function values. The principal branch is most common, giving $\ln(-1) = i\pi$ .

Practice Problems

Elementary Analytic Functions Practice

Questions

Correct

Accuracy

The complex exponential

e^z

is periodic with period:

Medium

Not attempted

The complex logarithm

\ln z

is:

Easy

Not attempted

For

z \in \mathbb{C}

\sin z

is:

Medium

Not attempted

The principal value of

\ln(-1)

is:

Medium

Not attempted

For complex

z

\cos z

and

\sin z

satisfy:

Easy

Not attempted

The function

z^{1/2}

(square root) has:

Hard

Not attempted

For

z = x + iy

, the real and imaginary parts of

e^z

are:

Easy

Not attempted

The branch cut for the principal branch of

\ln z

is typically taken along:

Medium

Not attempted

For

z \in \mathbb{C}

\sinh z

is related to

\sin z

by:

Medium

Not attempted

The function

z^i

(where

i

is the imaginary unit) is:

Hard

Not attempted

Frequently Asked Questions

Why is the complex exponential periodic with imaginary period?

Since $e^{z + 2\pi i} = e^z e^{2\pi i} = e^z \cdot 1 = e^z$, adding $2\pi i$ doesn't change the value. This is fundamentally different from real exponentials, which are not periodic.

What is a branch cut?

A branch cut is a curve in the complex plane along which a multi-valued function is made single-valued (single branch). It's typically taken along a ray or line where the function is discontinuous in its multi-valued form.

Why is the complex logarithm multi-valued?

The logarithm $\ln z = \ln|z| + i\arg z$ is multi-valued because the argument $\arg z$ can differ by any multiple of $2\pi$. Each value $\arg z + 2k\pi$ gives a different branch of the logarithm.

How do complex trigonometric functions differ from real ones?

Complex trigonometric functions are unbounded, while real ones are bounded. Also, identities like $\cos^2 z + \sin^2 z = 1$ still hold, but $|\sin z|$ and $|\cos z|$ can be arbitrarily large for complex arguments.

What is the principal value of a multi-valued function?

The principal value is the value chosen from among the many possible values, typically by restricting the argument to a specific range (like $(-\pi, \pi]$ for the logarithm) or choosing a specific branch.