Proof of Points of the Unit Circle and the Polar Form of a Nonzero Complex Number

Routine arithmetic in $\mathbb{R}$, $\mathbb{Z}$ and $\mathbb{C}$ is used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity). The existence part of (i) is proved by reaching the unit circle one quarter at a time: the closed first quadrant comes from the intermediate value theorem, and multiplication by $i$ and by $-1$, which are values of $\exp$, carries it to the rest of the circle.

The closed first quadrant. Let $z=a+bi$ with $a,b$ real, $a\ge0$, $b\ge0$ and $|z|=1$; we find $t\in[0,\tfrac{\pi}{2}]$ with $\exp(it)=z$. Here $a^{2}+b^{2}=|z|^{2}=1$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii)), so $a^{2}\le1$ and hence $0\le a\le1$. The cosine is continuous on the interval $[0,\tfrac{\pi}{2}]$, which has two distinct endpoints as $\pi>0$ (Quarter-Turn Identities and Periodicity of Sine and Cosine (i)): this is Continuity, Derivatives and Power Series of Sine and Cosine (i), restricted to the interval as in The Real Line: Standing Notation and Background for Calculus on Intervals (vi), intervals being as in The Real Line: Standing Notation and Background for Calculus on Intervals (v). Moreover

$$\cos\tfrac{\pi}{2}=0\le a\le1=\cos0$$

by Quarter-Turn Identities and Periodicity of Sine and Cosine (ii) and Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (i). The intermediate value theorem Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals (i) therefore gives $t\in[0,\tfrac{\pi}{2}]$ with $\cos t=a$. For this $t$ the sine is nonnegative, $\sin t\ge0$: if $t=0$, then $\sin t=0$, and if $0<t\le\tfrac{\pi}{2}$, then $t<\pi$ and $\sin t>0$ by Quarter-Turn Identities and Periodicity of Sine and Cosine (vii). The Pythagorean identity The Pythagorean Identity for Sine and Cosine (i) gives

$$(\sin t)^{2}=1-(\cos t)^{2}=1-a^{2}=b^{2},$$

and as $\sin t$ and $b$ are both nonnegative, $\sin t=b$. Hence, by Euler's formula Euler's Formula: the Complex Exponential on the Imaginary Axis, Its Modulus and Its Conjugate (i), $\exp(it)=\cos t+i\sin t=a+bi=z$.

The closed upper half. Let $|z|=1$ with $\operatorname{Im}z\ge0$; we find $t\in[0,\pi]$ with $\exp(it)=z$. If also $\operatorname{Re}z\ge0$, the first-quadrant case applies and even gives $t\in[0,\tfrac{\pi}{2}]$. If $\operatorname{Re}z<0$, we rotate $z$ back by a quarter turn: put $w=-iz$. Writing $z=a+bi$, we have $w=b-ai$, whose real part $b$ and imaginary part $-a$ are nonnegative; and

$$|w|=|{-i}|\,|z|=1,$$

by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv), since $|{-i}|=|i|=1$ by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii) (indeed $|i|\ge0$ and $|i|^{2}=0^{2}+1^{2}=1$). The first-quadrant case gives $s\in[0,\tfrac{\pi}{2}]$ with $\exp(is)=w$, and since $\exp\bigl(\tfrac{\pi}{2}i\bigr)=i$ by Periodicity of the Complex Exponential and Its Kernel (i), the functional equation The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (ii) gives

$$z=iw=\exp\bigl(\tfrac{\pi}{2}i\bigr)\exp(is)=\exp\bigl(i(s+\tfrac{\pi}{2})\bigr),\qquad s+\tfrac{\pi}{2}\in[\tfrac{\pi}{2},\pi].$$

So in either case $z=\exp(it)$ for some $t\in[0,\pi]$.

Existence in (i). Let $|z|=1$. If $\operatorname{Im}z\ge0$, the upper-half case gives $t\in[0,\pi]$ with $\exp(it)=z$. If $\operatorname{Im}z<0$, then $-z$ has positive imaginary part and modulus $|{-z}|=|z|=1$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii)), so the upper-half case gives $s\in[0,\pi]$ with $\exp(is)=-z$; since $\exp(\pi i)=-1$ by Periodicity of the Complex Exponential and Its Kernel (i), the functional equation The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (ii) gives

$$z=(-1)(-z)=\exp(\pi i)\exp(is)=\exp\bigl(i(s+\pi)\bigr),\qquad s+\pi\in[\pi,2\pi].$$

In every case, then, $z=\exp(it)$ for some $t$ with $0\le t\le2\pi$. The endpoint $t=2\pi$ is not allowed in (i), but if it occurs, then $z=\exp(2\pi i)=1=\exp(0)$ by Periodicity of the Complex Exponential and Its Kernel (i) and The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (i), and $t=0$ serves instead. So there is $t\in[0,2\pi)$ with $\exp(it)=z$.

Uniqueness in (i). Let $0\le s,t<2\pi$ with $\exp(it)=\exp(is)$. By The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (ii) and The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (iii),

$$\exp\bigl(i(t-s)\bigr)=\exp(it)\exp(-is)=\frac{\exp(it)}{\exp(is)}=1,$$

so the description of the kernel in Periodicity of the Complex Exponential and Its Kernel (iii) gives an integer $k$ with $i(t-s)=2\pi ki$; comparing imaginary parts (The Complex Numbers, with the Real Numbers Identified with a Subset of the Complex Numbers (vii)), $t-s=2\pi k$. Since $-2\pi<t-s<2\pi$ and $\pi>0$ (Quarter-Turn Identities and Periodicity of Sine and Cosine (i)), dividing by $2\pi$ gives $-1<k<1$. By the discreteness of the integers, Discreteness of the Integers (i), $-1<k$ gives $0\le k$ and $k<1$ gives $k\le0$; hence $k=0$ and $t=s$.

The polar form. Let $z\neq0$. Then $|z|>0$ by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii), and $u=z/|z|$ satisfies

$$|u|=|z|\,\bigl||z|^{-1}\bigr|=\frac{|z|}{|z|}=1,$$

by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv) and, as $|z|^{-1}$ is a positive real number, The Modulus of a Complex Number (ii). By (i) there is $t\in[0,2\pi)$ with $\exp(it)=u$, and then

$$z=|z|\,u=r\exp(it)\qquad\text{with } r=|z|>0 .$$

Conversely, suppose $z=r\exp(it)$ with $r>0$ and $t\in[0,2\pi)$. Then $|z|=r\,|\exp(it)|=r$, by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv), The Modulus of a Complex Number (ii) and Euler's Formula: the Complex Exponential on the Imaginary Axis, Its Modulus and Its Conjugate (ii); so necessarily $r=|z|$. Consequently $\exp(it)=z/|z|$, a point of the unit circle, and $t$ is determined by the uniqueness in (i). Hence the pair $(r,t)$ is unique, and $r=|z|$. This proves (ii).