Definition 1.1Definition of cis

다음과 같이 함수 cis:  R→C\text{cis:} \; \mathbb{R}\to\mathbb{C}를 정의한다.

cis(θ)=cos⁡θ+isin⁡θ\text{cis} (\theta) = \cos \theta + i \sin \theta

Thesis 1.2Thesis of magnitude of cis

다음이 성립한다.

∀z∈C,  ∣z∣=∣cis(θ)∣=1\forall z \in \mathbb{C}, \; |z| = |\text{cis} (\theta)| = 1

By the definition of cis,

∣cis(θ)∣=∣cos⁡θ+isin⁡θ∣=cos⁡2θ+sin⁡2θ=1|\text{cis} (\theta)| = |\cos \theta + i \sin \theta| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1■_\blacksquare

Thesis 1.3Thesis of complex polar form

다음이 성립한다.

∀z∈C∃!r∈R+θ∈(−π,π]  s.t.  z=r⋅cis(θ)\forall z \in \mathbb{C} \exists ! r \in \mathbb{R}^{+} \theta \in (-\pi,\pi] \; s.t. \; z = r \cdot \text{cis} (\theta)

By polar expression of complex number,

∀z∈C∃!r∈R+θ∈(−π,π]  s.t.  z=r⋅(cos⁡θ+isin⁡θ)\forall z \in \mathbb{C} \exists ! r \in \mathbb{R}^{+} \theta \in (-\pi,\pi] \; s.t. \; z = r \cdot ( \cos \theta + i \sin \theta )

=r⋅cis(θ)= r \cdot \text{cis} (\theta)(1.1)■_\blacksquare

Thesis 1.4Lemma: cis product/addition rule

다음이 성립한다.

∀α,β∈R    cis(α) cis(β)=cis(α+β)\forall \alpha,\beta \in \mathbb{R} \;\; \text{cis}(\alpha)\,\text{cis}(\beta) = \text{cis}(\alpha+\beta)

From the definition of cis\text{cis} and trig addition formulas,

cis(α) cis(β)=(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)\text{cis}(\alpha)\,\text{cis}(\beta) = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta)

=(cos⁡αcos⁡β−sin⁡αsin⁡β)+i(sin⁡αcos⁡β+cos⁡αsin⁡β)= (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)

=cos⁡(α+β)+isin⁡(α+β)=cis(α+β)= \cos(\alpha+\beta) + i\sin(\alpha+\beta) = \text{cis}(\alpha+\beta)(1.1)■_\blacksquare

Thesis 1.5De Moivre's theorem for cis

다음이 성립한다.

∀θ∈R  ∀n∈N,  (cis θ)n=cis(nθ)\forall \theta \in \mathbb{R}\; \forall n \in \mathbb{N}, \; (\text{cis}\,\theta)^n = \text{cis}(n\theta)

    Base n=0:  (cis θ)0=1=cis(0)=cis(0⋅θ).\;\;\text{Base } n=0:\; (\text{cis}\,\theta)^0 = 1 = \text{cis}(0) = \text{cis}(0\cdot\theta).

    Induction n→n+1:  (cis θ)n+1=(cis θ)n cis θ\;\;\text{Induction } n\to n+1:\; (\text{cis}\,\theta)^{n+1} = (\text{cis}\,\theta)^n\,\text{cis}\,\theta

        =cis(nθ) cis(θ)=cis((n+1)θ)\;\;\;\;= \text{cis}(n\theta)\,\text{cis}(\theta) = \text{cis}((n+1)\theta)(1.4)■_\blacksquare

FollowUp 1.5-1De Moivre's theorem for cis

이제 이로 부터 자명하게 복소수의 polar form에서의 성질들이 cis form에서도 성립함을 알 수 있다.

Definition 1.6Definition of arg

다음과 같이 함수 arg:  C∖{0}→(−π,π]\text{arg:} \; \mathbb{C}\setminus\{0\}\to (-\pi,\pi] 를 정의한다.

θ=arg(z)⇔θ∈(−π,π]∧∃r∈R+  s.t.  z=rcis(θ)\theta = \text{arg} (z) \Leftrightarrow \theta \in (-\pi,\pi] \land \exist r \in \R^{+} \; s.t. \; z=r \text{cis}(\theta)

Definition 1.7Definition of ω

다음과 같이 ωn:  C×Z∩[0,n)→C\omega _{n}: \; \mathbb{C} \times \mathbb{Z} \cap[0, n) \to \mathbb{C} (where n∈Nn \in \mathbb{N}) 를 정의한다.

ωn(z,k)=∣z∣1n⋅cis(arg(z)+2kπn)\omega _{n} (z,k) = |z|^\frac{1}{n} \cdot \text{cis} \left( \frac {\text{arg} (z)+ 2k\pi}{n}\right)

Thesis 1.8Property of ω

다음이 성립한다.

∀n∈N  ∀z∈C  ∀k∈Z∩[0,n)  {ωn(z,k)}n=z\forall n \in \mathbb{N} \; \forall z \in \mathbb{C} \; \forall k \in \mathbb{Z} \cap [0,n) \; \{\omega_{n} (z,k)\}^{n} = z

{ωn(z,k)}n\{\omega_{n} (z,k)\}^{n}

={∣z∣1n⋅cis(arg(z)+2kπn)}n= \{ |z|^\frac{1}{n} \cdot \text{cis} \left( \frac {\text{arg} (z)+ 2k\pi}{n} \right) \}^{n}(1.7)

=∣z∣nn⋅cis(arg(z)+2kπ)= |z|^\frac{n}{n} \cdot \text{cis} \left( {\text{arg} (z)+ 2k\pi} \right)(1.5)

=∣z∣⋅cis(arg(z))= |z| \cdot \text{cis} \left( {\text{arg} (z)} \right)

=z= z(1.6)(1.1)■_\blacksquare

Thesis 1.9Roots of complex numbers

다음이 성립한다.

zn=c⇒∃k∈N  s.t.  z=ωkz^n = c \Rightarrow \exists k \in \mathbb{N} \; s.t. \; z = \omega_k

Let z=r⋅cis(θ)z = r \cdot \text{cis} (\theta)(1.3)

Let c=∣c∣⋅cis(ϕ)c = |c| \cdot \text{cis} (\phi)(1.3)    ⋯(1)\; \; \cdots \text{(1)}

Then, we have:

zn=rn⋅cis(nθ)=cz^n = r^n \cdot \text{cis} (n\theta) = c    ⋯(2)\; \; \cdots \text{(2)}

As, z→∣z∣z \to |z| is a function,

∣rn⋅cis(nθ)∣=∣c∣|r^n \cdot \text{cis} (n\theta)| = |c|

⇒∣rn∣⋅∣cis(nθ)∣=∣c∣\Rightarrow |r^n| \cdot |\text{cis} (n\theta)| = |c|

⇒∣r∣n⋅1=∣c∣\Rightarrow |r|^n \cdot 1 = |c|(1.2)

⇒∣r∣⋅1=∣c∣1n\Rightarrow |r| \cdot 1 = |c|^{\frac{1}{n}}

⇒r=∣c∣1n\Rightarrow r = |c|^{\frac{1}{n}}(∵r>0)(\because r>0)

Also, from (2), we have:

zn=∣c∣nn⋅cis(nθ)=c=∣c∣⋅cis(ϕ)z^n = |c|^{\frac{n}{n}} \cdot \text{cis} (n\theta) = c = |c| \cdot \text{cis} (\phi)(∵(1))(\because (1))

⇒cis(nθ)=cis(ϕ)\Rightarrow \text{cis} (n\theta) = \text{cis} (\phi)

⇒nθ=ϕ+2kπ,  k∈Z\Rightarrow n\theta = \phi + 2k\pi, \; k \in \mathbb{Z}

⇒θ=ϕ+2kπn,  k∈Z\Rightarrow \theta = \frac{\phi + 2k\pi}{n}, \; k \in \mathbb{Z}

∴zn=c⇒∃k∈N  s.t.  z=ωn(z,k)\therefore z^n = c \Rightarrow \exists k \in \mathbb{N} \; s.t. \; z = \omega_n (z,k)■_\blacksquare

Thesis 1.10nth roots of complex numbers

복소수zz의 nn제곱근은 nn개이며, 이들은 ωn(z,k)\omega_n (z,k)꼴이다.

(1.9), (1.8)에 의해 자명하게 제곱근은 ωn(z,k)\omega_n (z,k)꼴이다.

또한, k=0,1,⋯ ,n−1k=0,1,\cdots,n-1에 대해 서로 다름을 보이자.

∀k1,k2∈{0,1,⋯ ,n−1}(k1≠k2)\forall k_1,k_2 \in \{0,1,\cdots,n-1\} \left( k_1 \neq k_2 \right)

arg(ωn(z,k1))≁arg(ωn(z,k2))\text{arg} (\omega_n (z,k_1)) \nsim \text{arg} (\omega_n (z,k_2))

⇔arg(z)+2k1πn≁arg(z)+2k2πn\Leftrightarrow \frac {\text{arg} (z)+ 2k_1\pi}{n} \nsim \frac {\text{arg} (z)+ 2k_2\pi}{n}

⇔2k1πn≁2k2πn\Leftrightarrow \frac {2k_1\pi}{n} \nsim \frac {2k_2\pi}{n}

otherwise, ∃p∈Z−{0}  s.t.  2k1πn∼2k2πn+2pπ  \exist p \in \mathbb{Z} - \{0\} \; s.t. \; \frac {2k_1\pi}{n} \sim \frac {2k_2\pi}{n}+2p\pi \;(∵k1≠k2)(\because k_1 \neq k_2)

⇔2(k1−k2−pn)π=0\Leftrightarrow 2(k_1-k_2- pn)\pi = 0

⇔k1−k2=pn\Leftrightarrow k_1-k_2= pn

⇔∣k1−k2∣=∣p∣n\Leftrightarrow |k_1-k_2| = |p|n    ⋯(*)\; \; \cdots \text{(*)}

since 0≤k1,k2≤n−10 \leq k_1,k_2 \leq n-1

−n+1≤k1−k2≤n−1-n+1 \leq k_1-k_2 \leq n-1

⇒∣k1−k2∣<n<∣p∣n  \Rightarrow |k_1 - k_2| < n < |p|n \;(∵p≠0)(\because p \neq 0)

⇒∣k1−k2∣≠∣p∣n\Rightarrow |k_1 - k_2| \neq |p|n

따라서, (*)에 모순되므로 주어진 명제는 참이다.■_\blacksquare