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✓ 21 results · all verified · 0 also independently AI-judged
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The Complex Exponential and Euler's Formula

1 · Prerequisites

2 · Summary

This development treats C as the Euclidean plane. The declared prerequisites provide real exponential and trigonometric series, metric completeness, finite monoid sums, absolute convergence, and compactness; Heine--Borel and the metric extreme-value theorem are used explicitly in the minimum-modulus argument.

It constructs complex arithmetic, series, exponential, logarithms, and trigonometric functions, then proves Euler's formula, the exponential fibre classification, polar form, roots of unity, and the principal-logarithm conventions. The final polynomial lemmas establish growth, global minimum modulus, and d'Alembert descent, yielding the fundamental theorem of algebra.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

Integer powers in the complex field

Definition

Fix z∈C. Apply The recursion theorem to the set C, the initial value 1, and the function w↦wz. This defines the natural powers uniquely by z0=1,zn+1=znz(n∈N).

Let ȷ ⁣:N→Z be the embedding of The naturals embed in the integers. For an integer r≥0, that lemma gives a unique n∈N with r=ȷ(n); define zr:=zn. If r<0 and z≠0, there is a unique n∈N with n≥1 and −r=ȷ(n); define zr:=(zn)−1. The inverse exists because C is a field by C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2). Thus nonnegative integer powers are defined for every z, while negative integer powers are defined exactly when z≠0; in particular 00=1 and no negative power of 0 is defined. The integer and its unique natural representative are never conflated.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The binomial theorem over the complex field

Statement

Facts & Assumptions

Given: Complex z,w and natural n.

Proof

technique · induction
1.1

For n=0 both sides are the empty-sum convention 1.

base
1.2

Assume the formula at n.

ih
2.1

Multiply the formula in step 1.2 by z+w, split the finite initial-segment sums, prove the shift k↦k+1 from the recursive monoid-sum clauses, and group equal powers.

given
3.1

Pascal's rule gives the complex coefficient ιC ⁣((n+1k)) at every index, including the endpoints, so the formula holds at n+1.

discharge-induction∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane

Definition

For z=x+iy and w=u+iv, put dC(z,w):=∣z−w∣=(x−u)2+(y−v)2=∥(x−u,y−v)∥2. Under the identification C=R2, this is exactly the metric d2 induced by the Euclidean norm of The p-norms ∥x∥p for rational p≥1, and ∥x∥∞. It is a metric by Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page, so the metric axioms are established rather than assumed.

Convergence in C, Cauchy sequences in C, and continuity of maps between subsets of C mean the notions of Convergence of a sequence in a metric space: xk→x iff d(xk,x)→0 in R, Cauchy sequence in a metric space, and Continuity of a map between metric spaces, at a point and globally, in the ε-δ form for dC (and the restricted metric on a subset). These uses are therefore licensed by the metric-space definition Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts

Statement

The metric space C is complete. A sequence zn=xn+iyn converges to x+iy exactly when xn→x and yn→y. The conventions and prerequisite facts used below are recorded in The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, For n≥1 a sequence in Rn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn is complete in every norm.

Facts & Assumptions

Given: A complex sequence zn=xn+iyn.

Proof

technique · direct
1.1

The complex metric is the Euclidean metric on R2.

given
2.1

Apply componentwise convergence and completeness in the published Euclidean-space theorem.

given∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

Complex series, absolute convergence, complex power series, and radius of convergence

Definition

A complex sequence is a function a ⁣:N→C. Since the additive reduct (C,+,0) of the complex field is a commutative monoid, the finite-list construction of The product g0g1⋯gn−1 of a finite list in a monoid, by recursion, with the empty product (n=0) equal to the identity, read additively, defines its complex partial sums by S0=0,SN+1=SN+aN,SN=∑n<NCan. The superscript is omitted when the summands already make the codomain clear.

The complex series ∑an converges to s∈C when SN→s in the metric of The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane; then s is its sum. This is the same finite sum and convergence as the vector-series definition on R2 in Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums. The series converges absolutely when the real series ∑∣an∣ of Series, partial sums, convergence and the sum, divergence, and the tail series converges. For a bijection σ ⁣:N→N, its rearrangement along σ is the complex series ∑aσ(n).

Let (cn) be a complex sequence and a∈C. The complex power series centered at a with coefficients (cn) is ∑n=0∞cn(z−a)n, with powers from Integer powers in the complex field and the preceding complex partial sums. To define its radius without ambiguity, form the specific real power series Q(x):=∑n=0∞∣cn∣xn centered at 0. The radius of convergence of the complex power series is, by definition, the radius of this Q in A real power series about a centre, its interval of convergence, and its radius in [0,+∞]. Thus the coefficient sequence, real variable, and center of the comparison series are all explicit.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Every absolutely convergent complex series converges, and rearrangements preserve its sum

Statement

Every absolutely convergent complex series converges, and every rearrangement has the same sum. The conventions and prerequisite facts used below are recorded in Complex series, absolute convergence, complex power series, and radius of convergence, An absolutely convergent series in Rn converges, and every rearrangement converges to the same sum.

Facts & Assumptions

Given: A complex series with convergent modulus series.

Proof

technique · direct
1.1

Regard its terms as vectors in R2; the Euclidean norm is the complex modulus.

given
2.1

Apply the absolute-convergence and rearrangement theorem for finite-dimensional vector series.

given∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Cauchy-Hadamard for complex power series, including zero and infinite radius

Statement

For ∑n≥0cn(z−a)n, set L:=lim sup⁡k→∞∣ck+1∣1/(k+1)∈[0,+∞], so no 0th root occurs, and set R:={+∞,L=0,1/L,0<L<+∞,0,L=+∞. Then the series converges absolutely for ∣z−a∣<R and diverges for ∣z−a∣>R; no assertion is made on ∣z−a∣=R. At z=a it converges to c0, including when R=0. The conventions and prerequisite facts used below are recorded in Complex series, absolute convergence, complex power series, and radius of convergence, Every absolutely convergent complex series converges, and rearrangements preserve its sum, Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾, For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently, Root test: lim sup⁡∣ak∣1/k<1 gives absolute convergence and hence convergence, >1 gives divergence, and =1 decides nothing.

Facts & Assumptions

Given: The coefficient sequence and a complex z.

[L1]

Root test: lim sup⁡∣ak∣1/k<1 gives absolute convergence and hence convergence, >1 gives divergence, and =1 decides nothing applies to a real series ∑n≥1an using the defined roots ∣ak+1∣1/(k+1).

[L2]

For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently states that if a real L is the limit superior of (bk), then bk>L−ε frequently for every real ε>0.

[L3]

Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾ defines lim sup⁡bk as the infimum of the extended-real tail suprema sup⁡{bj:j≥k}.

[L4]

Every absolutely convergent complex series converges, and rearrangements preserve its sum states that every absolutely convergent complex series converges.

[L5]

Every convergent sequence in a metric space is Cauchy states that every convergent sequence in a metric space is Cauchy.

[L6]

Complex series, absolute convergence, complex power series, and radius of convergence defines the partial sums by S0=0 and SN+1=SN+aN, and defines convergence through the complex metric.

Proof

technique · direct
1.1

At z=a, every positive-index term vanishes, so the series converges to c0.

algebra
1.2

Suppose z≠a, put r=∣z−a∣>0, and set bk=∣ck+1∣1/(k+1). For the real modulus tail dn=∣cn∣rn (n≥1), its root family is dk+11/(k+1)=bkr.

algebra
2.1

If r<R, then L is finite and L<1/r. Put ε=(1/r−L)/2>0. The eventual-upper-bound clause of [L2] gives bk<L+ε<1/r eventually; by [L3], the limit superior of the root family bkr is therefore at most (L+ε)r<1. Hence [L1] gives convergence of the modulus tail. (When L=+∞, R=0 and r<R is impossible.)

L1L2L3step 1.2algebra
2.2

Suppose 0<L<+∞ and r>R=1/L. Then ε:=L−1/r>0, and [L2] gives bk>L−ε=1/r frequently. At those arbitrarily large indices, step 1.2 gives dk+1=(bkr)k+1>1.

L2step 1.2algebra
2.3

Suppose L=+∞ and r>R=0. By [L3], every tail supremum of (bk) is +∞; hence 1/r is not an upper bound for any tail, so bk>1/r frequently. Again dk+1=(bkr)k+1>1 at arbitrarily large indices.

L3step 1.2algebra
3.1

In either divergence case, let SN be the complex partial sums. If (SN) converged, [L5] would make it Cauchy; but [L6] gives dC(Sn+1,Sn)=∣cn(z−a)n∣=dn>1 for arbitrarily large n, contradicting the Cauchy condition with tolerance 1. Thus the complex series diverges.

L5L6step 2.2step 2.3
3.2

In the case r<R, step 2.1 says that the complex series is absolutely convergent, so it converges by [L4].

L4step 2.1
4.1

Step 1.1 covers the centre, steps 3.1 and 3.2 cover respectively ∣z−a∣>R and ∣z−a∣<R, and none of these arguments asserts anything when 0<L<+∞ and ∣z−a∣=R. This proves all three radius cases exactly as stated.

step 1.1step 3.1step 3.2∎
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums

Statement

Facts & Assumptions

Given: Absolutely convergent complex series (an) and (bn).

Proof

technique · direct
1.1

Define cn by the recursive finite complex sum. Componentwise expansion is valid because complex addition and multiplication are coordinatewise polynomial formulas.

given
1.2

Put dn=∑k≤n∣ak∣∣bn−k∣. The triangle and multiplicative modulus laws give ∣cn∣≤dn.

algebra
1.3

The real absolute Cauchy-product theorem makes ∑dn converge; comparison therefore makes ∑∣cn∣ converge.

given
2.1

Expanding real and imaginary parts gives four real Cauchy products. Their sums combine by real series linearity to the two coordinates of AB, and componentwise convergence identifies ∑cn=AB.

given∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

The complex exponential by its power series

Definition

Let ιR ⁣:N→R be the canonical-natural map of The canonical natural ι(n)=n⋅1F of a field, and let j ⁣:R→C be the real embedding j(x)=x+i0 from The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i. For every n∈N, the factorial n! is nonzero by The factorial n! and the falling factorial nk‾, defined by recursion in N, so ιR(n!)>0 by Canonical naturals are positive and strictly increasing and consequently j(ιR(n!))≠0 in the complex field C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2).

For z∈C, define exp⁡z:=∑n=0∞znj(ιR(n!)). whenever this complex series converges. Inside a complex expression we abbreviate the embedded denominator j(ιR(n!)) by n!, so the same definition may be written exp⁡z=∑zn/n! without identifying a natural number with a complex number. Powers and series are those of Integer powers in the complex field and Complex series, absolute convergence, complex power series, and radius of convergence. The convergence for every z∈C is discharged by The complex exponential series converges absolutely for every complex argument ↗.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The complex exponential series converges absolutely for every complex argument

Statement

For every z∈C, the series ∑zn/n! converges absolutely. The conventions and prerequisite facts used below are recorded in The complex exponential by its power series, The exponential series converges absolutely for every real argument, Every absolutely convergent complex series converges, and rearrangements preserve its sum.

Facts & Assumptions

Given: z∈C.

Proof

technique · direct
1.1

Its modulus series is ∑∣z∣n/n!, the real exponential series at the nonnegative real ∣z∣.

algebra
2.1

The real infinite-radius lemma and complex absolute-convergence theorem give the result.

given∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential

Facts & Assumptions

Given: Complex z,w and real x.

[L1]

The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums says that the product has nth coefficient ∑k≤nakbn−k.

[L3]

The binomial theorem over the complex field gives (z+w)n=∑k≤nιC ⁣((nk))zkwn−k, where ιC is its canonical-natural map.

[L4]

The complex exponential series converges absolutely for every complex argument states that ∑zn/n! converges absolutely for every complex z.

Proof

technique · direct
1.1

By [L4], the two exponential series converge absolutely, so [L1] makes their product the Cauchy product.

L1L4
2.1

Its degree-n coefficient is ∑k≤nzkwn−k/(k!(n−k)!). By [L2], after applying the canonical-natural map into C, each summand is ιC ⁣((nk))zkwn−k/n!, and [L3] turns their sum into (z+w)n/n!.

L2L3step 1.1
3.1

The resulting series is the defining series of exp⁡(z+w). When z=x+0i, every term is the corresponding real term in the definition of ex, so the two values agree.

step 2.1∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Euler's formula: exp⁡(iθ)=cos⁡θ+isin⁡θ for every real θ

Statement

For every real θ, exp⁡(iθ)=cos⁡θ+isin⁡θ.

Facts & Assumptions

Given: A real θ.

[L1]

The complex exponential by its power series defines exp⁡z as the sum of the complex series with terms zn/n!, where the factorial is embedded in the complex field.

[L2]

The complex exponential series converges absolutely for every complex argument states that for every z∈C, the series ∑zn/n! converges absolutely.

[L3]
[L4]

Every absolutely convergent complex series converges, and rearrangements preserve its sum states that every absolutely convergent complex series converges.

[L6]

Convergent series add and scale termwise states that convergent real series add and scale termwise, with the corresponding sums.

[L8]

Sine and cosine defined by their real power series defines sine and cosine by the real series ∑k=0∞(−1)kθ2k+1/(2k+1)! and ∑k=0∞(−1)kθ2k/(2k)!.

[L9]

Integer powers in the complex field defines natural complex powers by z0=1 and zn+1=znz.

Proof

technique · direct
1.1

Put an=(iθ)n/n!, and define the parity masks by en=an for even n and en=0 for odd n, while on=0 for even n and on=an for odd n. By [L1] and [L2], ∑an converges absolutely and has sum exp⁡(iθ).

L1L2construct
1.2

Induction from i2=−1 and the recursion in [L9] gives i2k=(−1)k and i2k+1=i(−1)k for every k∈N. Hence e2k=(−1)kθ2k/(2k)!, e2k+1=0, o2k=0, and o2k+1=i(−1)kθ2k+1/(2k+1)!.

L9algebra
2.1

Since 0≤∣en∣≤∣an∣ and 0≤∣on∣≤∣an∣, [L3] shows that both modulus series converge. Thus ∑en and ∑on are absolutely convergent and hence converge by [L4].

L3L4step 1.1
3.1

By [L5], the two coordinate series of e and o converge. Since an=en+on, applying [L6] in each coordinate gives ∑n=0∞an=∑n=0∞en+∑n=0∞on.

L5L6step 2.1
3.2

Apply [L7] to each real coordinate series of e, with consecutive blocks [2k,2k+2). By step 1.2 the real-coordinate blocks are (−1)kθ2k/(2k)! and the imaginary-coordinate blocks are 0. Therefore [L8] identifies ∑en=cos⁡θ.

L5L7L8step 2.1step 1.2
3.3

Apply the same coordinatewise grouping to o. Its real-coordinate blocks are 0 and its imaginary-coordinate blocks are (−1)kθ2k+1/(2k+1)!, so [L8] gives ∑on=isin⁡θ.

L5L7L8step 2.1step 1.2
4.1

Substitute steps 3.2 and 3.3 into step 3.1 and use step 1.1: exp⁡(iθ)=cos⁡θ+isin⁡θ.

step 1.1step 3.1step 3.2step 3.3∎
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0

Statement

Facts & Assumptions

Given: Reals x,y.

Proof

technique · direct
1.1

Apply the addition law to x+iy and Euler's formula.

given
1.2

Multiplicativity of modulus, the Pythagorean identity, and positivity of ex give the modulus formula.

given
2.1

The defining quarter-turn value cos⁡π=−1 and sin⁡π=0 give Euler's identity.

given∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

ker⁡(exp⁡)=2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ

Facts & Assumptions

Given: z=x+iy and w=u+iv.

Proof

technique · direct
1.1

Cartesian form shows exp⁡z=1 forces ex=1 and (cos⁡y,sin⁡y)=(1,0).

given
1.2

Strict monotonicity gives x=0, while the trigonometric period theorem gives y∈2πZ.

given
2.1

The addition law turns equality of two exponential values into membership of z−w in the kernel, and the converse is immediate.

given∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Every nonzero complex number has a unique polar form r(cos⁡θ+isin⁡θ) with r>0 and −π<θ≤π

Statement

Every z≠0 has a unique representation z=r(cos⁡θ+isin⁡θ) with r=∣z∣>0 and −π<θ≤π. The conventions and prerequisite facts used below are recorded in Real and imaginary parts, complex conjugation, and modulus, t↦(cos⁡t,sin⁡t) is a bijection from [0,2π) onto the real unit circle, The zero sets of sine and cosine and the least positive common period 2 pi.

Facts & Assumptions

Given: z=x+iy≠0.

Proof

technique · direct
1.1

The point (x/∣z∣,y/∣z∣) lies on the unit circle.

algebra
1.2

The unit-circle parametrization supplies an angle, and its endpoint convention converts it uniquely to (−π,π].

given
2.1

Multiplying by ∣z∣ proves existence; the period theorem proves uniqueness.

given∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The complex exponential maps C onto C∖{0}

Facts & Assumptions

Given: z∈C∖{0}.

Proof

technique · constructive
1.1

Write z=r(cos⁡θ+isin⁡θ) with r>0.

given
1.2

Put w=log⁡r+iθ.

construct
2.1

Cartesian exponential form and elog⁡r=r give exp⁡w=z.

discharge-construct∎
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Complex de Moivre formula for every integer exponent

Statement

Let j:Z→Q be the integer embedding of The integers embed in the rationals, let ιR:Q→R be the ordered-field embedding of The unique embedding of ℚ into an ordered field, and put κR:=ιR∘j. For every integer m and real θ, (cos⁡θ+isin⁡θ)m=cos⁡(κR(m)θ)+isin⁡(κR(m)θ). The conventions and prerequisite facts used below are recorded in Euler's formula: exp⁡(iθ)=cos⁡θ+isin⁡θ for every real θ, exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2), Integer powers in the complex field.

Facts & Assumptions

Given: An integer m and real θ.

Proof

technique · direct
1.1

Euler's formula identifies the base with exp⁡(iθ).

given
1.2

Repeated addition handles nonnegative powers by the exponential addition law; inverses handle negative powers.

given
2.1

Euler's formula at κR(m)θ gives the displayed result.

given∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The n-th roots of a complex number and the n distinct roots of unity for every n≥1

Statement

Write ιR:N→R for the canonical-natural map of The canonical natural ι(n)=n⋅1F of a field. If n∈N with n≥1 and z=reiθ≠0, its n distinct roots are r1/nexp⁡ ⁣(iθ+2πιR(k)ιR(n)),k∈N,0≤k<n. For z=0, the only nth root is 0. Thus the nth roots of unity are precisely exp⁡ ⁣(i2πιR(k)ιR(n)) for k∈N with 0≤k<n. The conventions and prerequisite facts used below are recorded in Every nonzero complex number has a unique polar form r(cos⁡θ+isin⁡θ) with r>0 and −π<θ≤π, Complex de Moivre formula for every integer exponent, ker⁡(exp⁡)=2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ, Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, The canonical natural ι(n)=n⋅1F of a field, Integer powers in the complex field.

Facts & Assumptions

Given: n∈N with n≥1 and z∈C.

Proof

technique · constructive
1.1

For z≠0, construct each listed candidate from the positive real nth root of r; de Moivre verifies it.

construct
1.2

The kernel theorem shows two listed candidates coincide only when their indices are equal modulo n.

given
2.1

Conversely polar form and the same kernel calculation force every root onto the list; the z=0 case is immediate.

discharge-construct∎
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

For n≥2, the sum of all n-th roots of unity is zero

Facts & Assumptions

Given: A natural n≥2 and ζ=exp⁡ ⁣(i2πιR(n)), with ιR as in The n-th roots of a complex number and the n distinct roots of unity for every n≥1.

Proof

technique · direct
1.1

The roots are the distinct list 1,ζ,…,ζn−1, with ζn=1 and ζ≠1.

given
1.2

The cyclic successor map on the initial segment n is a permutation; the commutative-monoid permutation rule therefore gives ζS=S for S=∑k<nζk.

given
2.1

Thus (ζ−1)S=0, and field cancellation gives S=0.

algebra∎
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Complex logarithms, the principal logarithm, and principal and multivalued complex powers

Definition

For z≠0 with principal polar form z=r(cos⁡θ+isin⁡θ), −π<θ≤π, define the principal logarithm Log⁡z:=log⁡r+iθ. The set of all complex logarithms of z is log⁡multiz:={v∈C:exp⁡v=z}. For w∈C, define the principal power and the multivalued power respectively by zprw:=exp⁡(wLog⁡z),zmultiw:={exp⁡(wv):v∈log⁡multiz}. Thus the first is one specified complex number and the second is a set of complex numbers; neither notation silently identifies them. The conventions and prerequisite facts used below are recorded in Every nonzero complex number has a unique polar form r(cos⁡θ+isin⁡θ) with r>0 and −π<θ≤π, The natural logarithm as the inverse of the exponential function, The complex exponential by its power series, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2).

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

All logarithms of z≠0 are Log⁡z+2πik, k∈Z

Statement

For z≠0, the solutions of exp⁡w=z are exactly Log⁡z+2πik for k∈Z. The conventions and prerequisite facts used below are recorded in Complex logarithms, the principal logarithm, and principal and multivalued complex powers, ker⁡(exp⁡)=2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ, The complex exponential maps C onto C∖{0}.

Facts & Assumptions

Given: z≠0 and w∈C.

Proof

technique · direct
1.1

The principal logarithm is one solution by its polar definition.

given
2.1

Equality exp⁡w=exp⁡(Log⁡z) is equivalent to w−Log⁡z∈2πiZ by the fibre theorem.

given∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

There is no continuous logarithm on all of C∖{0}

Statement

Facts & Assumptions

Given: The unit-circle path γ(t)=exp⁡(it) for 0≤t≤2π.

[L1]

ker⁡(exp⁡)=2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ states that exp⁡u=exp⁡v exactly when u−v∈2πiZ.

[L2]

Euler's formula: exp⁡(iθ)=cos⁡θ+isin⁡θ for every real θ gives γ(t)=cos⁡t+isin⁡t, and The derivatives of sine and cosine are cosine and minus sine makes both real coordinate functions continuous.

[L5]

The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane defines continuity on subsets of C by its Euclidean metric.

Proof

technique · contradiction
1.1

Suppose such a continuous L exists and put h(t)=L(γ(t))−it. By [L2], [L3], and [L5], γ, L∘γ, and h are continuous on [0,2π].

assume-contraL2L3L5
2.1

The assumed identity says exp⁡(L(γ(t)))=γ(t)=exp⁡(it). Hence [L1] gives h(t)∈2πiZ for every t.

L1step 1.1
3.1

By [L3], ν(t):=Im⁡(h(t))/(2π) is a continuous real-valued function; by step 2.1 it takes values in Z. If ν(s)≠ν(t) for some s<t, [L4] applied to ν on [s,t] gives a noninteger value strictly between two distinct integers, a contradiction. Thus ν is constant.

L3L4step 2.1
4.1

Euler's formula gives γ(0)=γ(2π)=1, so h(2π)=L(1)−2πi=h(0)−2πi. Its imaginary quotient therefore changes by −1, contradicting step 3.1.

L2step 1.1step 3.1discharge-contradiction∎
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential

Definition

For z∈C, define sin⁡z=exp⁡(iz)−exp⁡(−iz)2i,cos⁡z=exp⁡(iz)+exp⁡(−iz)2, sinh⁡z=exp⁡z−exp⁡(−z)2,cosh⁡z=exp⁡z+exp⁡(−z)2. The conventions and prerequisite facts used below are recorded in The complex exponential by its power series, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2).

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The exponential formulas, real restrictions, and trigonometric-hyperbolic dictionary over C

Statement

For every z∈C, cos⁡z=cosh⁡(iz),sin⁡z=−isinh⁡(iz),cosh⁡z=cos⁡(iz),sinh⁡z=−isin⁡(iz). The complex functions restrict to their real sine, cosine, hyperbolic sine and hyperbolic cosine series on the real axis. The conventions and prerequisite facts used below are recorded in Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential, Euler's formula: exp⁡(iθ)=cos⁡θ+isin⁡θ for every real θ, exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential.

Facts & Assumptions

Given: A complex number z.

Proof

technique · direct
1.1

Substitute iz and −iz into the four definitions and simplify i2=−1.

algebra
2.1

On real arguments, group the exponential series into even and odd terms as in Euler's formula.

given∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

Formal complex polynomials, evaluation, degree, leading coefficient, and monic polynomials

Definition

A formal complex polynomial is either the zero polynomial 0, or a finite coefficient list (a0,…,an) with an≠0; we write the latter as p(Z)=∑k=0nakZk. The list, rather than the function it induces, is the polynomial object. Define evaluation at z∈C by 0(z):=0,p(z):=∑k<n+1Cakzkfor p=(a0,…,an)≠0. where the latter is the initial-segment complex sum defined in Complex series, absolute convergence, complex power series, and radius of convergence. Thus evaluation is defined for the zero polynomial as well as every nonzero formal polynomial.

For nonzero p=(a0,…,an), define deg⁡p:=n and lc⁡(p):=an. The zero polynomial has no degree and no leading coefficient. A nonzero polynomial is monic when lc⁡(p)=1. Complex arithmetic and powers are those of C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2) and Integer powers in the complex field.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A nonconstant complex polynomial tends to infinite modulus and attains a global minimum modulus

Statement

Facts & Assumptions

Given: A nonconstant polynomial p(z)=anzn+∑k<nakzk with an≠0.

[L1]

Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive gives ∣uv∣=∣u∣∣v∣ and ∣u+v∣≤∣u∣+∣v∣.

[L3]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value gives a minimum for a continuous real-valued function on a nonempty compact metric space.

Proof

technique · direct
1.1

Put C:=∑k<n∣ak∣. For r=∣z∣≥1, [L1] gives ∣p(z)∣≥∣an∣rn−∑k<n∣ak∣rk≥rn(∣an∣−C/r). Thus for r≥2C/∣an∣ the right side is at least (∣an∣/2)rn, which tends to +∞.

L1algebra
1.2

Writing z=x+iy, each coordinate projection is continuous because ∣x−x0∣,∣y−y0∣≤∥(x,y)−(x0,y0)∥. The identity uv−u0v0=u(v−v0)+v0(u−u0) proves continuity of products, so induction over the finite expression makes both coordinate polynomials of p(x+iy) continuous. Then [L4] makes p and ∣p∣ continuous.

L4
2.1

Choose R≥1 so that the lower bound of step 1.1 is >∣p(0)∣ when ∣z∣>R. The closed square K=[−R,R]2⊆R2=C is nonempty and compact by [L2]; outside K one has ∣z∣>R. By [L3] and step 1.2, let a∈K minimize ∣p∣ on K.

L2L3step 1.1step 1.2
3.1

Since 0∈K, this minimizer satisfies ∣p(a)∣≤∣p(0)∣. Step 2.1 makes every point outside K have strictly larger modulus, so a is a global minimizer.

step 2.1∎
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A nonzero value of a nonconstant complex polynomial cannot be a local minimum of its modulus

Facts & Assumptions

Given: A nonconstant polynomial p and a point a with p(a)≠0.

[L1]

The n-th roots of a complex number and the n distinct roots of unity for every n≥1 supplies an mth root of every nonzero complex number when m≥1.

[L2]

Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive gives ∣uv∣=∣u∣∣v∣, the triangle inequality, and vv‾=∣v∣2.

[L3]

The binomial theorem over the complex field gives the finite expansion of (a+z)n in complex coefficients.

Proof

technique · constructive
1.1

By [L3], expanding p(a+z)−p(a) gives a nonzero polynomial in z (its top coefficient is the nonzero leading coefficient of p). Let m≥1 be its first nonzero degree, so p(a+z)=p(a)+cmzm+zm+1q(z) with cm≠0. Choose by [L1] a unit complex number u with um=−p(a)cm‾/(∣p(a)∣∣cm∣). Then p(a)‾cmum=−∣p(a)∣∣cm∣.

L1L3constructalgebra
1.2

Write M for the sum of the moduli of the finitely many coefficients of q. For 0<t≤1, [L2] gives ∣tm+1q(tu)∣≤Mtm+1. With λ:=∣p(a)∣∣cm∣ and B:=2∣p(a)∣M+(∣cm∣+M)2>0, choose 0<t<min⁡{1,λ/B}.

L2choose
2.1

Put R=tm+1q(tu). By step 1.1 and [L2], ∣p(a+tu)∣2≤∣p(a)∣2−2λtm+2∣p(a)∣Mtm+1+(∣cm∣+M)2t2m≤∣p(a)∣2−2λtm+Btm+1<∣p(a)∣2. Thus a+tu is arbitrarily close to a and has strictly smaller modulus.

L2step 1.1step 1.2discharge-construct∎
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Fundamental theorem of algebra: every nonconstant complex polynomial has a complex root

Statement

Every nonconstant complex polynomial has a complex root. The conventions and prerequisite facts used below are recorded in A nonconstant complex polynomial tends to infinite modulus and attains a global minimum modulus, A nonzero value of a nonconstant complex polynomial cannot be a local minimum of its modulus.

Facts & Assumptions

Given: A nonconstant complex polynomial p.

[L2]

A nonzero value of a nonconstant complex polynomial cannot be a local minimum of its modulus states that, if p(a)≠0, then ∣p∣ is not minimal on any neighbourhood of a.

Proof

technique · contradiction
1.1

By [L1], let a minimize ∣p∣ globally.

L1
1.2

Suppose p(a)≠0.

assume-contra
2.1

By [L2], there is a point arbitrarily near a with strictly smaller modulus, contradicting step 1.1.

L2step 1.1
3.1

Therefore p(a)=0, proving the theorem.

discharge-contradiction∎

5 · Examples, counterexamples and false statements

None yet.

Sources