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23 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 23 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Complex Exponential and Euler's Formula

1 · Prerequisites

2 · Summary

This development treats C\mathbb 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: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

The complex numbers as R2\mathbb R^2, with their arithmetic, real embedding, and imaginary unit

Definition

Set C=R2\mathbb C=\mathbb R^2, write (x,y)=x+iy(x,y)=x+iy, embed xRx\in\mathbb R as (x,0)(x,0), and put i=(0,1)i=(0,1). Define (x+iy)+(u+iv)=(x+u)+i(y+v),(x+iy)(u+iv)=(xuyv)+i(xv+yu).(x+iy)+(u+iv)=(x+u)+i(y+v),\qquad (x+iy)(u+iv)=(xu-yv)+i(xv+yu). Then i2=1i^2=-1. The conventions and prerequisite facts used below are recorded in The real numbers.

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

The complex numbers form a field, and every nonzero x+iyx+iy has inverse (xiy)/(x2+y2)(x-iy)/(x^2+y^2)

Statement

C\mathbb C with the preceding operations is a field. For z=x+iy0z=x+iy\ne0, z1=xiyx2+y2.z^{-1}=\frac{x-iy}{x^2+y^2}. The conventions and prerequisite facts used below are recorded in The complex numbers as R2\mathbb R^2, with their arithmetic, real embedding, and imaginary unit, The reals form a totally ordered field, Field.

Facts & Assumptions

Given: Complex numbers z=x+iyz=x+iy and w=u+ivw=u+iv.

Proof

technique · direct
1.1

Coordinate expansion verifies associativity, commutativity, distributivity, and the identities 00 and 11.

algebra
1.2

If z0z\ne0, then x2+y2>0x^2+y^2>0 and direct multiplication gives z(xiy)=x2+y2z(x-iy)=x^2+y^2.

algebra
2.1

Dividing by this nonzero real proves the inverse formula and the field axioms.

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

Integer powers in the complex field

Definition

Fix zCz\in\mathbb C. Apply The recursion theorem to the set C\mathbb C, the initial value 11, and the function wwzw\mapsto wz. This defines the natural powers uniquely by

z0=1,zn+1=znz(nN).z^0=1,\qquad z^{n+1}=z^nz\quad(n\in\mathbb N).

Let ȷ ⁣:NZ\jmath\colon\mathbb N\to\mathbb Z be the embedding of The naturals embed in the integers. For an integer r0r\ge0, that lemma gives a unique nNn\in\mathbb N with r=ȷ(n)r=\jmath(n); define zr:=znz^r:=z^n. If r<0r<0 and z0z\ne0, there is a unique nNn\in\mathbb N with n1n\ge1 and r=ȷ(n)-r=\jmath(n); define

zr:=(zn)1.z^r:=(z^n)^{-1}.

The inverse exists because C\mathbb C is a field by The complex numbers form a field, and every nonzero x+iyx+iy has inverse (xiy)/(x2+y2)(x-iy)/(x^2+y^2). Thus nonnegative integer powers are defined for every zz, while negative integer powers are defined exactly when z0z\ne0; in particular 00=10^0=1 and no negative power of 00 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,wz,w and natural nn.

Proof

technique · induction
1.1

For n=0n=0 both sides are the empty-sum convention 11.

base
1.2

Assume the formula at nn.

ih
2.1

Multiply the formula in step 1.2 by z+wz+w, split the finite initial-segment sums, prove the shift kk+1k\mapsto 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))\iota_{\mathbb C}\!\left(\binom{n+1}k\right) at every index, including the endpoints, so the formula holds at n+1n+1.

discharge-induction
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Real and imaginary parts, complex conjugation, and modulus

Definition

For z=x+iyz=x+iy, set Rez=x\operatorname{Re}z=x, Imz=y\operatorname{Im}z=y, z=xiy\overline z=x-iy, and z=x2+y20|z|=\sqrt{x^2+y^2}\ge0. The conventions and prerequisite facts used below are recorded in The complex numbers as R2\mathbb R^2, with their arithmetic, real embedding, and imaginary unit, Square roots exist: a unique a0\sqrt{a} \ge 0 with (a)2=a(\sqrt{a})^2 = a; the positives are {x2:x0}\{x^2 : x \neq 0\}.

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

Conjugation laws, zz=z2z\overline z=|z|^2, multiplicativity of modulus, and the triangle inequality

Statement

Facts & Assumptions

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

Proof

technique · direct
1.1

Expand the coordinate definitions to prove the conjugation laws and zz=x2+y2=z2z\overline z=x^2+y^2=|z|^2.

algebra
1.2

Squaring both nonnegative sides proves multiplicativity of the modulus.

algebra
2.1

The Euclidean norm triangle inequality on R2\mathbb R^2 is exactly z+wz+w|z+w|\le|z|+|w|.

given
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+iyz=x+iy and w=u+ivw=u+iv, put

dC(z,w):=zw=(xu)2+(yv)2=(xu,yv)2.d_{\mathbb C}(z,w):=|z-w|=\sqrt{(x-u)^2+(y-v)^2}=\lVert(x-u,y-v)\rVert_2.

Under the identification C=R2\mathbb C=\mathbb R^2, this is exactly the metric d2d_2 induced by the Euclidean norm of The pp-norms xp\lVert x\rVert_p for rational p1p \ge 1, and x\lVert x\rVert_\infty. It is a metric by Each p\lVert\cdot\rVert_p is a norm on Rn\mathbb{R}^n, and the induced metrics are exactly d1d_1, d2d_2 and dd_\infty of the published metric-spaces page, so the metric axioms are established rather than assumed.

Convergence in C\mathbb C, Cauchy sequences in C\mathbb C, and continuity of maps between subsets of C\mathbb C mean the notions of Convergence of a sequence in a metric space: xkxx_k \to x iff d(xk,x)0d(x_k, x) \to 0 in R\mathbb{R}, Cauchy sequence in a metric space, and Continuity of a map between metric spaces, at a point and globally, in the ε\varepsilon-δ\delta form for dCd_{\mathbb C} (and the restricted metric on a subset). These uses are therefore licensed by the metric-space definition Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = 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\mathbb C is complete. A sequence zn=xn+iynz_n=x_n+iy_n converges to x+iyx+iy exactly when xnxx_n\to x and ynyy_n\to y. The conventions and prerequisite facts used below are recorded in The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, For n1n \ge 1 a sequence in Rn\mathbb{R}^n converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn\mathbb{R}^n is complete in every norm.

Facts & Assumptions

Given: A complex sequence zn=xn+iynz_n=x_n+iy_n.

Proof

technique · direct
1.1

The complex metric is the Euclidean metric on R2\mathbb R^2.

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 ⁣:NCa\colon\mathbb N\to\mathbb C. Since the additive reduct (C,+,0)(\mathbb C,+,0) of the complex field is a commutative monoid, the finite-list construction of The product g0g1gn1g_0 g_1 \cdots g_{n-1} of a finite list in a monoid, by recursion, with the empty product (n=0n = 0) equal to the identity, read additively, defines its complex partial sums by

S0=0,SN+1=SN+aN,SN=n<NCan.S_0=0,\qquad S_{N+1}=S_N+a_N,\qquad S_N=\sum_{n<N}^{\mathbb C}a_n.

The superscript is omitted when the summands already make the codomain clear.

The complex series an\sum a_n converges to sCs\in\mathbb C when SNsS_N\to s in the metric of The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane; then ss is its sum. This is the same finite sum and convergence as the vector-series definition on R2\mathbb R^2 in Series of vectors in Rn\mathbb{R}^n, absolute convergence, rearrangement, and the set of rearrangement sums. The series converges absolutely when the real series an\sum |a_n| of Series, partial sums, convergence and the sum, divergence, and the tail series converges. For a bijection σ ⁣:NN\sigma\colon\mathbb N\to\mathbb N, its rearrangement along σ\sigma is the complex series aσ(n)\sum a_{\sigma(n)}.

Let (cn)(c_n) be a complex sequence and aCa\in\mathbb C. The complex power series centered at aa with coefficients (cn)(c_n) is

n=0cn(za)n,\sum_{n=0}^{\infty}c_n(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=0cnxnQ(x):=\sum_{n=0}^{\infty}|c_n|x^n

centered at 00. The radius of convergence of the complex power series is, by definition, the radius of this QQ in A real power series about a centre, its interval of convergence, and its radius in [0,+][0,+\infty]. 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\mathbb{R}^n 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\mathbb R^2; 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 n0cn(za)n\sum_{n\ge0}c_n(z-a)^n, set L:=lim supkck+11/(k+1)[0,+],L:=\limsup_{k\to\infty}|c_{k+1}|^{1/(k+1)}\in[0,+\infty], so no 00th root occurs, and set R:={+,L=0,1/L,0<L<+,0,L=+.R:=\begin{cases}+\infty,&L=0,\\1/L,&0<L<+\infty,\\0,&L=+\infty.\end{cases} Then the series converges absolutely for za<R|z-a|<R and diverges for za>R|z-a|>R; no assertion is made on za=R|z-a|=R. At z=az=a it converges to c0c_0, including when R=0R=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 infnsupknxk\inf_n \sup_{k \ge n} x_k and supninfknxk\sup_n \inf_{k \ge n} x_k in R\overline{\mathbb{R}}, For finite LL: L=lim supxkL = \limsup x_k iff for every ε>0\varepsilon > 0 one has xk<L+εx_k < L + \varepsilon eventually and xk>Lεx_k > L - \varepsilon frequently, Root test: lim supak1/k<1\limsup |a_k|^{1/k} < 1 gives absolute convergence and hence convergence, >1> 1 gives divergence, and =1= 1 decides nothing.

Facts & Assumptions

Given: The coefficient sequence and a complex zz.

[L1]

Root test: lim supak1/k<1\limsup |a_k|^{1/k} < 1 gives absolute convergence and hence convergence, >1> 1 gives divergence, and =1= 1 decides nothing applies to a real series n1an\sum_{n\ge1}a_n using the defined roots ak+11/(k+1)|a_{k+1}|^{1/(k+1)}.

[L2]

For finite LL: L=lim supxkL = \limsup x_k iff for every ε>0\varepsilon > 0 one has xk<L+εx_k < L + \varepsilon eventually and xk>Lεx_k > L - \varepsilon frequently states that if a real LL is the limit superior of (bk)(b_k), then bk>Lεb_k>L-\varepsilon frequently for every real ε>0\varepsilon>0.

[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=0S_0=0 and SN+1=SN+aNS_{N+1}=S_N+a_N, and defines convergence through the complex metric.

Proof

technique · direct
1.1

At z=az=a, every positive-index term vanishes, so the series converges to c0c_0.

algebra
1.2

Suppose zaz\ne a, put r=za>0r=|z-a|>0, and set bk=ck+11/(k+1)b_k=|c_{k+1}|^{1/(k+1)}. For the real modulus tail dn=cnrnd_n=|c_n|r^n (n1)(n\ge1), its root family is dk+11/(k+1)=bkrd_{k+1}^{1/(k+1)}=b_kr.

algebra
2.1

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

L1L2L3step 1.2algebra
2.2

Suppose 0<L<+0<L<+\infty and r>R=1/Lr>R=1/L. Then ε:=L1/r>0\varepsilon:=L-1/r>0, and [L2] gives bk>Lε=1/rb_k>L-\varepsilon=1/r frequently. At those arbitrarily large indices, step 1.2 gives dk+1=(bkr)k+1>1d_{k+1}=(b_kr)^{k+1}>1.

L2step 1.2algebra
2.3

Suppose L=+L=+\infty and r>R=0r>R=0. By [L3], every tail supremum of (bk)(b_k) is ++\infty; hence 1/r1/r is not an upper bound for any tail, so bk>1/rb_k>1/r frequently. Again dk+1=(bkr)k+1>1d_{k+1}=(b_kr)^{k+1}>1 at arbitrarily large indices.

L3step 1.2algebra
3.1

In either divergence case, let SNS_N be the complex partial sums. If (SN)(S_N) converged, [L5] would make it Cauchy; but [L6] gives dC(Sn+1,Sn)=cn(za)n=dn>1d_{\mathbb C}(S_{n+1},S_n)=|c_n(z-a)^n|=d_n>1 for arbitrarily large nn, contradicting the Cauchy condition with tolerance 11. Thus the complex series diverges.

L5L6step 2.2step 2.3
3.2

In the case r<Rr<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 za>R|z-a|>R and za<R|z-a|<R, and none of these arguments asserts anything when 0<L<+0<L<+\infty and za=R|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)(a_n) and (bn)(b_n).

Proof

technique · direct
1.1

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

given
1.2

Put dn=knakbnkd_n=\sum_{k\le n}|a_k||b_{n-k}|. The triangle and multiplicative modulus laws give cndn|c_n|\le d_n.

algebra
1.3

The real absolute Cauchy-product theorem makes dn\sum d_n converge; comparison therefore makes cn\sum|c_n| 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 ABAB, and componentwise convergence identifies cn=AB\sum c_n=AB.

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

The complex exponential by its power series

Definition

Let ιR ⁣:NR\iota_{\mathbb R}\colon\mathbb N\to\mathbb R be the canonical-natural map of The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, and let j ⁣:RCj\colon\mathbb R\to\mathbb C be the real embedding j(x)=x+i0j(x)=x+i0 from The complex numbers as R2\mathbb R^2, with their arithmetic, real embedding, and imaginary unit. For every nNn\in\mathbb N, the factorial n!n! is nonzero by The factorial n!n! and the falling factorial nkn^{\underline{k}}, defined by recursion in N\mathbb{N}, so ιR(n!)>0\iota_{\mathbb R}(n!)>0 by Canonical naturals are positive and strictly increasing and consequently j(ιR(n!))0j(\iota_{\mathbb R}(n!))\ne0 in the complex field The complex numbers form a field, and every nonzero x+iyx+iy has inverse (xiy)/(x2+y2)(x-iy)/(x^2+y^2).

For zCz\in\mathbb C, define

expz:=n=0znj(ιR(n!)).\exp z:=\sum_{n=0}^{\infty}\frac{z^n}{j(\iota_{\mathbb R}(n!))}.

whenever this complex series converges. Inside a complex expression we abbreviate the embedded denominator j(ιR(n!))j(\iota_{\mathbb R}(n!)) by n!n!, so the same definition may be written expz=zn/n!\exp z=\sum z^n/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 zCz\in\mathbb 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 zCz\in\mathbb C, the series zn/n!\sum z^n/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: zCz\in\mathbb C.

Proof

technique · direct
1.1

Its modulus series is zn/n!\sum |z|^n/n!, the real exponential series at the nonnegative real z|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)=expzexpw\exp(z+w)=\exp z\,\exp w, and the complex exponential extends the real exponential

Statement

Facts & Assumptions

Given: Complex z,wz,w and real xx.

[L1]

The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums says that the product has nnth coefficient knakbnk\sum_{k\le n}a_kb_{n-k}.

[L3]

The binomial theorem over the complex field gives (z+w)n=knιC ⁣((nk))zkwnk(z+w)^n=\sum_{k\le n}\iota_{\mathbb C}\!\left(\binom nk\right)z^kw^{n-k}, where ιC\iota_{\mathbb C} is its canonical-natural map.

[L4]

The complex exponential series converges absolutely for every complex argument states that zn/n!\sum z^n/n! converges absolutely for every complex zz.

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-nn coefficient is knzkwnk/(k!(nk)!)\sum_{k\le n}z^kw^{n-k}/(k!(n-k)!). By [L2], after applying the canonical-natural map into C\mathbb C, each summand is ιC ⁣((nk))zkwnk/n!\iota_{\mathbb C}\!\left(\binom nk\right)z^kw^{n-k}/n!, and [L3] turns their sum into (z+w)n/n!(z+w)^n/n!.

L2L3step 1.1
3.1

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

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

Euler's formula: exp(iθ)=cosθ+isinθ\exp(i\theta)=\cos\theta+i\sin\theta for every real θ\theta

Statement

For every real θ\theta, exp(iθ)=cosθ+isinθ\exp(i\theta)=\cos\theta+i\sin\theta.

Facts & Assumptions

Given: A real θ\theta.

[L1]

The complex exponential by its power series defines expz\exp z as the sum of the complex series with terms zn/n!z^n/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 zCz\in\mathbb C, the series zn/n!\sum z^n/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)!\sum_{k=0}^{\infty}(-1)^k\theta^{2k+1}/(2k+1)! and k=0(1)kθ2k/(2k)!\sum_{k=0}^{\infty}(-1)^k\theta^{2k}/(2k)!.

[L9]

Integer powers in the complex field defines natural complex powers by z0=1z^0=1 and zn+1=znzz^{n+1}=z^nz.

Proof

technique · direct
1.1

Put an=(iθ)n/n!a_n=(i\theta)^n/n!, and define the parity masks by en=ane_n=a_n for even nn and en=0e_n=0 for odd nn, while on=0o_n=0 for even nn and on=ano_n=a_n for odd nn. By [L1] and [L2], an\sum a_n converges absolutely and has sum exp(iθ)\exp(i\theta).

L1L2construct
1.2

Induction from i2=1i^2=-1 and the recursion in [L9] gives i2k=(1)ki^{2k}=(-1)^k and i2k+1=i(1)ki^{2k+1}=i(-1)^k for every kNk\in\mathbb N. Hence e2k=(1)kθ2k/(2k)!e_{2k}=(-1)^k\theta^{2k}/(2k)!, e2k+1=0e_{2k+1}=0, o2k=0o_{2k}=0, and o2k+1=i(1)kθ2k+1/(2k+1)!o_{2k+1}=i(-1)^k\theta^{2k+1}/(2k+1)!.

L9algebra
2.1

Since 0enan0\le|e_n|\le|a_n| and 0onan0\le|o_n|\le|a_n|, [L3] shows that both modulus series converge. Thus en\sum e_n and on\sum o_n are absolutely convergent and hence converge by [L4].

L3L4step 1.1
3.1

By [L5], the two coordinate series of ee and oo converge. Since an=en+ona_n=e_n+o_n, applying [L6] in each coordinate gives n=0an=n=0en+n=0on\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}e_n+\sum_{n=0}^{\infty}o_n.

L5L6step 2.1
3.2

Apply [L7] to each real coordinate series of ee, with consecutive blocks [2k,2k+2)[2k,2k+2). By step 1.2 the real-coordinate blocks are (1)kθ2k/(2k)!(-1)^k\theta^{2k}/(2k)! and the imaginary-coordinate blocks are 00. Therefore [L8] identifies en=cosθ\sum e_n=\cos\theta.

L5L7L8step 2.1step 1.2
3.3

Apply the same coordinatewise grouping to oo. Its real-coordinate blocks are 00 and its imaginary-coordinate blocks are (1)kθ2k+1/(2k+1)!(-1)^k\theta^{2k+1}/(2k+1)!, so [L8] gives on=isinθ\sum o_n=i\sin\theta.

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)=cosθ+isinθ\exp(i\theta)=\cos\theta+i\sin\theta.

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

exp(x+iy)=ex(cosy+isiny)\exp(x+iy)=e^x(\cos y+i\sin y), exp(x+iy)=ex|\exp(x+iy)|=e^x, and eiπ+1=0e^{i\pi}+1=0

Statement

Facts & Assumptions

Given: Reals x,yx,y.

Proof

technique · direct
1.1

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

given
1.2

Multiplicativity of modulus, the Pythagorean identity, and positivity of exe^x give the modulus formula.

given
2.1

The defining quarter-turn value cosπ=1\cos\pi=-1 and sinπ=0\sin\pi=0 give Euler's identity.

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

ker(exp)=2πiZ\ker(\exp)=2\pi i\mathbb Z, and expz=expw\exp z=\exp w exactly when zw2πiZz-w\in2\pi i\mathbb Z

Facts & Assumptions

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

Proof

technique · direct
1.1

Cartesian form shows expz=1\exp z=1 forces ex=1e^x=1 and (cosy,siny)=(1,0)(\cos y,\sin y)=(1,0).

given
1.2

Strict monotonicity gives x=0x=0, while the trigonometric period theorem gives y2πZy\in2\pi\mathbb Z.

given
2.1

The addition law turns equality of two exponential values into membership of zwz-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(cosθ+isinθ)r(\cos\theta+i\sin\theta) with r>0r>0 and π<θπ-\pi<\theta\le\pi

Statement

Every z0z\ne0 has a unique representation z=r(cosθ+isinθ)z=r(\cos\theta+i\sin\theta) with r=z>0r=|z|>0 and π<θπ-\pi<\theta\le\pi. The conventions and prerequisite facts used below are recorded in Real and imaginary parts, complex conjugation, and modulus, t(cost,sint)t\mapsto(\cos t,\sin t) is a bijection from [0,2π)[0,2\pi) 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+iy0z=x+iy\ne0.

Proof

technique · direct
1.1

The point (x/z,y/z)(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](-\pi,\pi].

given
2.1

Multiplying by z|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\mathbb C onto C{0}\mathbb C\setminus\{0\}

Facts & Assumptions

Given: zC{0}z\in\mathbb C\setminus\{0\}.

Proof

technique · constructive
1.1

Write z=r(cosθ+isinθ)z=r(\cos\theta+i\sin\theta) with r>0r>0.

given
1.2

Put w=logr+iθw=\log r+i\theta.

construct
2.1

Cartesian exponential form and elogr=re^{\log r}=r give expw=z\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:ZQj:\mathbb Z\to\mathbb Q be the integer embedding of The integers embed in the rationals, let ιR:QR\iota_{\mathbb R}:\mathbb Q\to\mathbb R be the ordered-field embedding of The unique embedding of ℚ into an ordered field, and put κR:=ιRj\kappa_{\mathbb R}:=\iota_{\mathbb R}\circ j. For every integer mm and real θ\theta, (cosθ+isinθ)m=cos(κR(m)θ)+isin(κR(m)θ).(\cos\theta+i\sin\theta)^m=\cos(\kappa_{\mathbb R}(m)\theta)+i\sin(\kappa_{\mathbb R}(m)\theta). The conventions and prerequisite facts used below are recorded in Euler's formula: exp(iθ)=cosθ+isinθ\exp(i\theta)=\cos\theta+i\sin\theta for every real θ\theta, exp(z+w)=expzexpw\exp(z+w)=\exp z\,\exp w, and the complex exponential extends the real exponential, The complex numbers form a field, and every nonzero x+iyx+iy has inverse (xiy)/(x2+y2)(x-iy)/(x^2+y^2), Integer powers in the complex field.

Facts & Assumptions

Given: An integer mm and real θ\theta.

Proof

technique · direct
1.1

Euler's formula identifies the base with exp(iθ)\exp(i\theta).

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\kappa_{\mathbb R}(m)\theta gives the displayed result.

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

The nn-th roots of a complex number and the nn distinct roots of unity for every n1n\ge1

Statement

Write ιR:NR\iota_{\mathbb R}:\mathbb N\to\mathbb R for the canonical-natural map of The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field. If nNn\in\mathbb N with n1n\ge1 and z=reiθ0z=re^{i\theta}\ne0, its nn distinct roots are r1/nexp ⁣(iθ+2πιR(k)ιR(n)),kN,0k<n.r^{1/n}\exp\!\left(i\frac{\theta+2\pi\iota_{\mathbb R}(k)}{\iota_{\mathbb R}(n)}\right),\qquad k\in\mathbb N,\quad 0\le k<n. For z=0z=0, the only nnth root is 00. Thus the nnth roots of unity are precisely exp ⁣(i2πιR(k)ιR(n))\exp\!\left(i\frac{2\pi\iota_{\mathbb R}(k)}{\iota_{\mathbb R}(n)}\right) for kNk\in\mathbb N with 0k<n0\le k<n. The conventions and prerequisite facts used below are recorded in Every nonzero complex number has a unique polar form r(cosθ+isinθ)r(\cos\theta+i\sin\theta) with r>0r>0 and π<θπ-\pi<\theta\le\pi, Complex de Moivre formula for every integer exponent, ker(exp)=2πiZ\ker(\exp)=2\pi i\mathbb Z, and expz=expw\exp z=\exp w exactly when zw2πiZz-w\in2\pi i\mathbb Z, Existence and uniqueness of nn-th roots: a unique a1/n0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Integer powers in the complex field.

Facts & Assumptions

Given: nNn\in\mathbb N with n1n\ge1 and zCz\in\mathbb C.

Proof

technique · constructive
1.1

For z0z\ne0, construct each listed candidate from the positive real nnth root of rr; de Moivre verifies it.

construct
1.2

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

given
2.1

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

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

For n2n\ge2, the sum of all nn-th roots of unity is zero

Facts & Assumptions

Given: A natural n2n\ge2 and ζ=exp ⁣(i2πιR(n))\zeta=\exp\!\left(i\frac{2\pi}{\iota_{\mathbb R}(n)}\right), with ιR\iota_{\mathbb R} as in The nn-th roots of a complex number and the nn distinct roots of unity for every n1n\ge1.

Proof

technique · direct
1.1

The roots are the distinct list 1,ζ,,ζn11,\zeta,\ldots,\zeta^{n-1}, with ζn=1\zeta^n=1 and ζ1\zeta\ne1.

given
1.2

The cyclic successor map on the initial segment nn is a permutation; the commutative-monoid permutation rule therefore gives ζS=S\zeta S=S for S=k<nζkS=\sum_{k<n}\zeta^k.

given
2.1

Thus (ζ1)S=0(\zeta-1)S=0, and field cancellation gives S=0S=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 z0z\ne0 with principal polar form z=r(cosθ+isinθ)z=r(\cos\theta+i\sin\theta), π<θπ-\pi<\theta\le\pi, define the principal logarithm Logz:=logr+iθ.\operatorname{Log}z:=\log r+i\theta. The set of all complex logarithms of zz is logmultiz:={vC:expv=z}.\log_{\mathrm{multi}}z:=\{v\in\mathbb C:\exp v=z\}. For wCw\in\mathbb C, define the principal power and the multivalued power respectively by zprw:=exp(wLogz),zmultiw:={exp(wv):vlogmultiz}.z^w_{\mathrm{pr}}:=\exp(w\operatorname{Log}z),\qquad z^w_{\mathrm{multi}}:=\{\exp(wv):v\in\log_{\mathrm{multi}}z\}. 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(cosθ+isinθ)r(\cos\theta+i\sin\theta) with r>0r>0 and π<θπ-\pi<\theta\le\pi, The natural logarithm as the inverse of the exponential function, The complex exponential by its power series, The complex numbers form a field, and every nonzero x+iyx+iy has inverse (xiy)/(x2+y2)(x-iy)/(x^2+y^2).

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

All logarithms of z0z\ne0 are Logz+2πik\operatorname{Log}z+2\pi i k, kZk\in\mathbb Z

Statement

For z0z\ne0, the solutions of expw=z\exp w=z are exactly Logz+2πik\operatorname{Log}z+2\pi ik for kZk\in\mathbb 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\ker(\exp)=2\pi i\mathbb Z, and expz=expw\exp z=\exp w exactly when zw2πiZz-w\in2\pi i\mathbb Z, The complex exponential maps C\mathbb C onto C{0}\mathbb C\setminus\{0\}.

Facts & Assumptions

Given: z0z\ne0 and wCw\in\mathbb C.

Proof

technique · direct
1.1

The principal logarithm is one solution by its polar definition.

given
2.1

Equality expw=exp(Logz)\exp w=\exp(\operatorname{Log}z) is equivalent to wLogz2πiZw-\operatorname{Log}z\in2\pi i\mathbb Z 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}\mathbb C\setminus\{0\}

Statement

Facts & Assumptions

Given: The unit-circle path γ(t)=exp(it)\gamma(t)=\exp(it) for 0t2π0\le t\le2\pi.

[L1]

ker(exp)=2πiZ\ker(\exp)=2\pi i\mathbb Z, and expz=expw\exp z=\exp w exactly when zw2πiZz-w\in2\pi i\mathbb Z states that expu=expv\exp u=\exp v exactly when uv2πiZu-v\in2\pi i\mathbb Z.

[L2]
[L5]

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

Proof

technique · contradiction
1.1

Suppose such a continuous LL exists and put h(t)=L(γ(t))ith(t)=L(\gamma(t))-it. By [L2], [L3], and [L5], γ\gamma, LγL\circ\gamma, and hh are continuous on [0,2π][0,2\pi].

assume-contraL2L3L5
2.1

The assumed identity says exp(L(γ(t)))=γ(t)=exp(it)\exp(L(\gamma(t)))=\gamma(t)=\exp(it). Hence [L1] gives h(t)2πiZh(t)\in2\pi i\mathbb Z for every tt.

L1step 1.1
3.1

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

L3L4step 2.1
4.1

Euler's formula gives γ(0)=γ(2π)=1\gamma(0)=\gamma(2\pi)=1, so h(2π)=L(1)2πi=h(0)2πih(2\pi)=L(1)-2\pi i=h(0)-2\pi i. Its imaginary quotient therefore changes by 1-1, 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 zCz\in\mathbb C, define sinz=exp(iz)exp(iz)2i,cosz=exp(iz)+exp(iz)2,\sin z=\frac{\exp(iz)-\exp(-iz)}{2i},\quad \cos z=\frac{\exp(iz)+\exp(-iz)}2, sinhz=expzexp(z)2,coshz=expz+exp(z)2.\sinh z=\frac{\exp z-\exp(-z)}2,\quad \cosh z=\frac{\exp z+\exp(-z)}2. The conventions and prerequisite facts used below are recorded in The complex exponential by its power series, The complex numbers form a field, and every nonzero x+iyx+iy has inverse (xiy)/(x2+y2)(x-iy)/(x^2+y^2).

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

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

Statement

For every zCz\in\mathbb C, cosz=cosh(iz),sinz=isinh(iz),coshz=cos(iz),sinhz=isin(iz).\cos z=\cosh(iz),\qquad \sin z=-i\sinh(iz),\qquad \cosh z=\cos(iz),\qquad \sinh z=-i\sin(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)=cosθ+isinθ\exp(i\theta)=\cos\theta+i\sin\theta for every real θ\theta, exp(z+w)=expzexpw\exp(z+w)=\exp z\,\exp w, and the complex exponential extends the real exponential.

Facts & Assumptions

Given: A complex number zz.

Proof

technique · direct
1.1

Substitute iziz and iz-iz into the four definitions and simplify i2=1i^2=-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 00, or a finite coefficient list (a0,,an)(a_0,\ldots,a_n) with an0a_n\ne0; we write the latter as p(Z)=k=0nakZkp(Z)=\sum_{k=0}^{n}a_kZ^k. The list, rather than the function it induces, is the polynomial object. Define evaluation at zCz\in\mathbb C by

0(z):=0,p(z):=k<n+1Cakzkfor p=(a0,,an)0.0(z):=0,\qquad p(z):=\sum_{k<n+1}^{\mathbb C}a_kz^k\quad\text{for }p=(a_0,\ldots,a_n)\ne0.

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)p=(a_0,\ldots,a_n), define degp:=n\deg p:=n and lc(p):=an\operatorname{lc}(p):=a_n. The zero polynomial has no degree and no leading coefficient. A nonzero polynomial is monic when lc(p)=1\operatorname{lc}(p)=1. Complex arithmetic and powers are those of The complex numbers form a field, and every nonzero x+iyx+iy has inverse (xiy)/(x2+y2)(x-iy)/(x^2+y^2) 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<nakzkp(z)=a_nz^n+\sum_{k<n}a_kz^k with an0a_n\ne0.

[L1]

Conjugation laws, zz=z2z\overline z=|z|^2, multiplicativity of modulus, and the triangle inequality gives uv=uv|uv|=|u||v| and u+vu+v|u+v|\le|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<nakC:=\sum_{k<n}|a_k|. For r=z1r=|z|\ge1, [L1] gives p(z)anrnk<nakrkrn(anC/r).|p(z)|\ge |a_n|r^n-\sum_{k<n}|a_k|r^k\ge r^n\bigl(|a_n|-C/r\bigr). Thus for r2C/anr\ge2C/|a_n| the right side is at least (an/2)rn(|a_n|/2)r^n, which tends to ++\infty.

L1algebra
1.2

Writing z=x+iyz=x+iy, each coordinate projection is continuous because xx0,yy0(x,y)(x0,y0)|x-x_0|,|y-y_0|\le\|(x,y)-(x_0,y_0)\|. The identity uvu0v0=u(vv0)+v0(uu0)uv-u_0v_0=u(v-v_0)+v_0(u-u_0) proves continuity of products, so induction over the finite expression makes both coordinate polynomials of p(x+iy)p(x+iy) continuous. Then [L4] makes pp and p|p| continuous.

L4
2.1

Choose R1R\ge1 so that the lower bound of step 1.1 is >p(0)>|p(0)| when z>R|z|>R. The closed square K=[R,R]2R2=CK=[-R,R]^2\subseteq\mathbb R^2=\mathbb C is nonempty and compact by [L2]; outside KK one has z>R|z|>R. By [L3] and step 1.2, let aKa\in K minimize p|p| on KK.

L2L3step 1.1step 1.2
3.1

Since 0K0\in K, this minimizer satisfies p(a)p(0)|p(a)|\le|p(0)|. Step 2.1 makes every point outside KK have strictly larger modulus, so aa 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 pp and a point aa with p(a)0p(a)\ne0.

[L1]

The nn-th roots of a complex number and the nn distinct roots of unity for every n1n\ge1 supplies an mmth root of every nonzero complex number when m1m\ge1.

[L2]

Conjugation laws, zz=z2z\overline z=|z|^2, multiplicativity of modulus, and the triangle inequality gives uv=uv|uv|=|u||v|, the triangle inequality, and vv=v2v\overline v=|v|^2.

[L3]

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

Proof

technique · constructive
1.1

By [L3], expanding p(a+z)p(a)p(a+z)-p(a) gives a nonzero polynomial in zz (its top coefficient is the nonzero leading coefficient of pp). Let m1m\ge1 be its first nonzero degree, so p(a+z)=p(a)+cmzm+zm+1q(z)p(a+z)=p(a)+c_mz^m+z^{m+1}q(z) with cm0c_m\ne0. Choose by [L1] a unit complex number uu with um=p(a)cm/(p(a)cm)u^m=-p(a)\overline{c_m}/(|p(a)||c_m|). Then p(a)cmum=p(a)cm\overline{p(a)}c_mu^m=-|p(a)||c_m|.

L1L3constructalgebra
1.2

Write MM for the sum of the moduli of the finitely many coefficients of qq. For 0<t10<t\le1, [L2] gives tm+1q(tu)Mtm+1|t^{m+1}q(tu)|\le Mt^{m+1}. With λ:=p(a)cm\lambda:=|p(a)||c_m| and B:=2p(a)M+(cm+M)2>0B:=2|p(a)|M+(|c_m|+M)^2>0, choose 0<t<min{1,λ/B}0<t<\min\{1,\lambda/B\}.

L2choose
2.1

Put R=tm+1q(tu)R=t^{m+1}q(tu). By step 1.1 and [L2], p(a+tu)2p(a)22λtm+2p(a)Mtm+1+(cm+M)2t2mp(a)22λtm+Btm+1<p(a)2|p(a+tu)|^2\le |p(a)|^2-2\lambda t^m+2|p(a)|Mt^{m+1}+(|c_m|+M)^2t^{2m}\le |p(a)|^2-2\lambda t^m+Bt^{m+1}<|p(a)|^2. Thus a+tua+tu is arbitrarily close to aa 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 pp.

[L1]

A nonconstant complex polynomial tends to infinite modulus and attains a global minimum modulus states that p|p| attains a global minimum on C\mathbb C.

[L2]

A nonzero value of a nonconstant complex polynomial cannot be a local minimum of its modulus states that, if p(a)0p(a)\ne0, then p|p| is not minimal on any neighbourhood of aa.

Proof

technique · contradiction
1.1

By [L1], let aa minimize p|p| globally.

L1
1.2

Suppose p(a)0p(a)\ne0.

assume-contra
2.1

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

L2step 1.1
3.1

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

discharge-contradiction

5 · Examples, counterexamples and false statements

None yet.

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