Proof of The Complex Plane as $\mathbb{R}^2$: the Map $(x,y)\mapsto x+iy$ Is a Real-Linear Isometric Bijection

Routine facts about real and complex numbers and about maps are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity); the earlier and the current versions of the definitions of a metric space and of an open set agree (The Earlier and the New Versions of the Definitions of Metric Spaces, Convergence, Continuity and Compactness Agree (i), The Earlier and the New Versions of the Definitions of Metric Spaces, Convergence, Continuity and Compactness Agree (v)), so openness in either version may be tested by the same ball criterion.

Step 1: the bijection. We prove (i). By The Complex Numbers, with the Real Numbers Identified with a Subset of the Complex Numbers (vii), carried by Complex Analysis: Standing Notation (i), every $z\in\mathbb{C}$ can be written as $z=x+iy$ with $x,y$ real in exactly one way, and the two real numbers are the real and imaginary parts, $x=\operatorname{Re}z$ and $y=\operatorname{Im}z$ (Real and Imaginary Parts of a Complex Number (i)). Put $\Psi(z)=(\operatorname{Re}z,\operatorname{Im}z)$ for $z\in\mathbb{C}$. Then $\Phi(\Psi(z))=\operatorname{Re}z+i\operatorname{Im}z=z$ for every $z\in\mathbb{C}$. Conversely, for $p=(x,y)\in\mathbb{R}^{2}$ the number $\Phi(p)=x+iy$ is already in canonical form, so uniqueness of that form gives

$$\operatorname{Re}\Phi(x,y)=x,\qquad \operatorname{Im}\Phi(x,y)=y .$$

Thus $\Psi(\Phi(p))$ and $p$ have the same components, and a point of $\mathbb{R}^{2}$ is determined by its components (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iii)), so $\Psi(\Phi(p))=p$. Hence $\Phi$ is a bijection of $\mathbb{R}^{2}$ onto $\mathbb{C}$ with inverse $\Psi$, which is the stated formula for $\Phi^{-1}$.

Step 2: real linearity. We prove (ii). Let $p=(a,b)$ and $q=(c,d)$ be points of $\mathbb{R}^{2}$ and let $t\in\mathbb{R}$. The operations of $\mathbb{R}^{2}$ are componentwise (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv)), so $p+q=(a+c,b+d)$ and $tp=(ta,tb)$. On the other side, the arithmetic of the canonical form (The Complex Numbers, with the Real Numbers Identified with a Subset of the Complex Numbers (vii)), with the real number $t$ read as $t+0i$, gives

$$\Phi(p)+\Phi(q)=(a+bi)+(c+di)=(a+c)+(b+d)i=\Phi(p+q),$$ $$t\,\Phi(p)=(t+0i)(a+bi)=ta+(tb)i=\Phi(tp).$$

Step 3: the isometry. We prove (iii). Let $p,q\in\mathbb{R}^{2}$ and write $h=p-q=(h_{1},h_{2})$. Since $p=q+h$, Step 2 gives $\Phi(p)=\Phi(q)+\Phi(h)$, that is, $\Phi(p)-\Phi(q)=\Phi(p-q)$. By Step 1, $\operatorname{Re}\Phi(h)=h_{1}$ and $\operatorname{Im}\Phi(h)=h_{2}$, so the definition of the modulus (The Modulus of a Complex Number (i)) gives $|\Phi(h)|=\sqrt{h_{1}^{2}+h_{2}^{2}}$. The Euclidean norm of $h$ (The Euclidean Norm on $\mathbb{R}^n$ (i)) is the same number, since its square is the sum of the squares of the components (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i)). Finally the Euclidean distance is the norm of the difference (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii)). Altogether

$$|\Phi(p)-\Phi(q)|=|\Phi(h)|=\sqrt{h_{1}^{2}+h_{2}^{2}}=\lVert p-q\rVert=d_{E}(p,q).$$

Step 4: open sets. We prove (iv). Let $W\subseteq\mathbb{C}$ and put $V=\{p\in\mathbb{R}^{2}:\Phi(p)\in W\}$. First, balls correspond to balls: for every $p\in\mathbb{R}^{2}$ and every real $r>0$,

$$\Phi\bigl(B_{d_{E}}(p,r)\bigr)=B_{d_{\mathbb{C}}}(\Phi(p),r)$$

(Open Ball in a Metric Space (i)). Indeed, by Step 3 a point $p'\in\mathbb{R}^{2}$ satisfies $d_{E}(p,p')<r$ exactly when $|\Phi(p)-\Phi(p')|<r$, and by Step 1 every point of $\mathbb{C}$ is $\Phi(p')$ for exactly one $p'$. Consequently, $B_{d_{E}}(p,r)\subseteq V$ holds exactly when $B_{d_{\mathbb{C}}}(\Phi(p),r)\subseteq W$. Both directions now follow from the ball criterion for openness: in $(\mathbb{C},d_{\mathbb{C}})$ that of Open Subset of a Metric Space (i), and in $(\mathbb{R}^{2},d_{E})$ that of Open Subset of a Metric Space (i), the two definitions agreeing by The Earlier and the New Versions of the Definitions of Metric Spaces, Convergence, Continuity and Compactness Agree (v).

Suppose $W$ is open, and let $p\in V$. Then $\Phi(p)\in W$, so some ball $B_{d_{\mathbb{C}}}(\Phi(p),r)$ with $r>0$ lies in $W$, and therefore $B_{d_{E}}(p,r)\subseteq V$. Hence $V$ is open.

Conversely, suppose $V$ is open, and let $z\in W$. By Step 1, $z=\Phi(p)$ for some $p\in\mathbb{R}^{2}$, and $p\in V$. So some ball $B_{d_{E}}(p,r)$ with $r>0$ lies in $V$, and therefore $B_{d_{\mathbb{C}}}(z,r)\subseteq W$. Hence $W$ is open.