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The Complex Exponential and Euler's Formula: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

All values of $i^{i}$ are the positive real numbers $e^{- π / 2 - 2 π k}$ , $k \in Z$

Example

All values of $i^{i}$ are $e^{- π / 2 - 2 π k}$ for $k \in Z$ , and each is a positive real number. The conventions and prerequisite facts used below are recorded in Complex logarithms, the principal logarithm, and principal and multivalued complex powers, All logarithms of $z \neq 0$ are $Log z + 2 π i k$ , $k \in Z$ .

Facts & Assumptions

Given: $Log i = i π / 2$ .

Verification

1.1

The logarithms of $i$ are $i (π / 2 + 2 π k)$ .

given

2.1

Multiplication by $i$ gives the real exponents $- π / 2 - 2 π k$ , whose exponentials are positive.

algebra∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The logarithms of $- 1$ are $(2 k + 1) π i$ , $k \in Z$

Example

The logarithms of $- 1$ are $(2 k + 1) π i$ for $k \in Z$ . The conventions and prerequisite facts used below are recorded in All logarithms of $z \neq 0$ are $Log z + 2 π i k$ , $k \in Z$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ .

Facts & Assumptions

Given: $Log (- 1) = i π$ .

Verification

1.1

Add the kernel $2 π i Z$ to the principal logarithm.

given

2.1

Simplifying $i π + 2 π i k$ gives $(2 k + 1) π i$ for every integer $k$ .

algebra∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The five fifth roots of unity and their sum

Example

The fifth roots of unity are $\exp ⁣ (i \frac{2 π ι_{R} (k)}{ι_{R} (5)})$ for $k \in N$ with $0 \leq k < 5$ , and their sum is $0$ . The conventions and prerequisite facts used below are recorded in The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ , For $n \geq 2$ , the sum of all $n$ -th roots of unity is zero.

Facts & Assumptions

Given: $n = 5$ .

Verification

1.1

The root classification lists the five values $\exp ⁣ (i \frac{2 π ι_{R} (k)}{ι_{R} (5)})$ for $k \in N$ with $0 \leq k < 5$ .

given

2.1

The root-sum corollary gives that their sum is $0$ .

given∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Complex sine is unbounded on the imaginary axis

Example

For real $t$ , $\sin (i t) = i \sinh t$ , so complex sine is unbounded on the imaginary axis. The conventions and prerequisite facts used below are recorded in The exponential formulas, real restrictions, and trigonometric-hyperbolic dictionary over $C$ , The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ .

Facts & Assumptions

Given: A real parameter $t$ .

Verification

1.1

The dictionary gives $\sin (i t) = i \sinh t$ .

given

2.1

Since $\sinh t = (e^{t} - e^{- t}) / 2$ is unbounded as $t \to \infty$ , so is its modulus.

given∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

$((- 1)^{2})^{1 / 2} = 1 \neq - 1 = (- 1)^{2 (1 / 2)}$ for principal complex powers

Statement refuted

The principal-power law $(z^{a})^{b} = z^{a b}$ is false without branch hypotheses. The conventions and prerequisite facts used below are recorded in Complex logarithms, the principal logarithm, and principal and multivalued complex powers, All logarithms of $z \neq 0$ are $Log z + 2 π i k$ , $k \in Z$ .

Facts & Assumptions

Given: $z = - 1$ , $a = 2$ , and $b = 1 / 2$ .

Counterexample

1.1

The principal square of $- 1$ is $1$ , so $((- 1)^{2})^{1 / 2} = 1$ .

given

2.1

The principal value of $(- 1)^{1}$ is $- 1$ , so the two sides differ.

given∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

$1 = e^{2 π i}$ does not imply $0 = 2 π i$ : logarithms invert the exponential only modulo its kernel

Example

Although $1 = \exp (2 π i)$ , one may not conclude $0 = 2 π i$ : logarithms classify fibres only modulo $2 π i Z$ . The conventions and prerequisite facts used below are recorded in $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , All logarithms of $z \neq 0$ are $Log z + 2 π i k$ , $k \in Z$ .

Facts & Assumptions

Given: The two complex numbers $0$ and $2 π i$ .

Verification

1.1

Both lie in the fibre of $1$ , and their difference is the nonzero kernel element $2 π i$ .

given

2.1

Thus equality of exponential values identifies logarithms only modulo the kernel, not as equal complex numbers.

algebra∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

$f (x + i y) = e^{x} (\cos 2 y + i \sin 2 y)$ is continuous, satisfies $f (z + w) = f (z) f (w)$ and $f (1) = e$ , but is not the standard complex exponential

Statement refuted

Continuity, $f (z + w) = f (z) f (w)$ , and $f (1) = e$ do not characterize the standard complex exponential. The conventions and prerequisite facts used below are recorded in $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, The addition formulas for sine and cosine, The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi, The exponential function is strictly increasing, A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions.

Facts & Assumptions

Given: $f (x + i y) = e^{x} (\cos 2 y + i \sin 2 y)$ .

[L1]

The addition formulas for sine and cosine gives the sine and cosine formulas for every pair of real arguments.

[L2]

The exponential function is strictly increasing states that $x \mapsto e^{x}$ is continuous, and The derivatives of sine and cosine are cosine and minus sine makes sine and cosine continuous.

[L3]

A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions makes a complex-valued map continuous when its real and imaginary parts are continuous.

[L4]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential restricts to the real exponential law $e^{x + s} = e^{x} e^{s}$ .

[L5]

Quarter-turn values and shifts by pi/2 and pi gives $\sin (2 π) = 0$ and $\cos (2 π) = 1$ .

[L6]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ gives $e^{i π} + 1 = 0$ .

Counterexample

1.1

For $z = x + i y$ and $w = s + i t$ , [L4] and [L1] give $f (z + w) = e^{x + s} (\cos (2 y + 2 t) + i \sin (2 y + 2 t)) = f (z) f (w)$ , and $f (1) = e$ .

L1L4algebra

1.2

The coordinate maps $(x, y) \mapsto x$ and $(x, y) \mapsto 2 y$ are continuous by the Euclidean norm estimate. Composition with the continuous real functions in [L2] is continuous, and the identity $u v - u_{0} v_{0} = u (v - v_{0}) + v_{0} (u - u_{0})$ proves continuity of their products. Thus [L3] makes $f$ continuous.

L2L3

2.1

By [L5], $f (i π) = 1$ , while [L6] gives $\exp (i π) = - 1$ . Thus $f$ is not the standard exponential.

L5L6∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The complex geometric power series has radius $1$ and sums to $1 / (1 - z)$ for $∣ z ∣ < 1$

Example

For $∣ z ∣ < 1$ , the complex geometric series $\sum_{n \geq 0} z^{n}$ has radius $1$ and sum $1 / (1 - z)$ . The conventions and prerequisite facts used below are recorded in Cauchy-Hadamard for complex power series, including zero and infinite radius, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ , Every absolutely convergent complex series converges, and rearrangements preserve its sum, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions.

Facts & Assumptions

Given: A complex $z$ with $∣ z ∣ < 1$ .

[L1]

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges says that the real series $\sum_{n \geq 0} r^{n}$ converges for $∣ r ∣ < 1$ .

[L2]

Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive gives $∣ z^{n} ∣ = ∣ z ∣^{n}$ .

[L3]

Every absolutely convergent complex series converges, and rearrangements preserve its sum says that an absolutely convergent complex series converges.

[L4]

For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ gives $∣ z ∣^{n} \to 0$ when $∣ z ∣ < 1$ .

[L5]

Cauchy-Hadamard for complex power series, including zero and infinite radius defines the shifted coefficient limsup and its radius cases.

Verification

1.1

The shifted coefficient roots in [L5] are all $1$ , so it gives radius $1$ .

2.1

By [L2], the modulus series is the real geometric series $\sum ∣ z ∣^{n}$ , which converges by [L1]; therefore the complex series converges by [L3], say to $S$ . The finite identity $(1 - z) \sum_{k < n} z^{k} = 1 - z^{n}$ holds in the complex field. Since $∣ z^{n} ∣ = ∣ z ∣^{n} \to 0$ by [L2] and [L4], passing to the limit gives $(1 - z) S = 1$ ; since $z \neq 1$ , $S = 1 / (1 - z)$ .

L1L2L3L4∎

Sources

J. Lebl, Basic Analysis I: Complex Numbers and the Complex Exponential

The Complex Exponential and Euler's Formula: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

All values of iiii are the positive real numbers e−π/2−2πke−π/2−2πk, k∈Zk∈Z

Example

Facts & Assumptions

Verification

The logarithms of −1−1 are (2k+1)πi(2k+1)πi, k∈Zk∈Z

Example

Facts & Assumptions

Verification

The five fifth roots of unity and their sum

Example

Facts & Assumptions

Verification

Complex sine is unbounded on the imaginary axis

Example

Facts & Assumptions

Verification

((−1)2)1/2=1≠−1=(−1)2(1/2)((−1)2)1/2=1=−1=(−1)2(1/2) for principal complex powers

Statement refuted

Facts & Assumptions

Counterexample

1=e2πi1=e2πi does not imply 0=2πi0=2πi: logarithms invert the exponential only modulo its kernel

Example

Facts & Assumptions

Verification

f(x+iy)=ex(cos⁡2y+isin⁡2y)f(x+iy)=ex(cos2y+isin2y) is continuous, satisfies f(z+w)=f(z)f(w)f(z+w)=f(z)f(w) and f(1)=ef(1)=e, but is not the standard complex exponential

Statement refuted

Facts & Assumptions

Counterexample

The complex geometric power series has radius 11 and sums to 1/(1−z)1/(1−z) for ∣z∣<1∣z∣<1

Example

Facts & Assumptions

Verification

Sources

All values of $i^{i}$ are the positive real numbers $e^{- π / 2 - 2 π k}$ , $k \in Z$

The logarithms of $- 1$ are $(2 k + 1) π i$ , $k \in Z$

$((- 1)^{2})^{1 / 2} = 1 \neq - 1 = (- 1)^{2 (1 / 2)}$ for principal complex powers

$1 = e^{2 π i}$ does not imply $0 = 2 π i$ : logarithms invert the exponential only modulo its kernel

$f (x + i y) = e^{x} (\cos 2 y + i \sin 2 y)$ is continuous, satisfies $f (z + w) = f (z) f (w)$ and $f (1) = e$ , but is not the standard complex exponential

The complex geometric power series has radius $1$ and sums to $1 / (1 - z)$ for $∣ z ∣ < 1$