Understanding Needham's derivation of equivalence between symbolic and geometric rules for complex multiplication Tristan Needham's Visual Complex Analysis contains a proof that the symbolic (i.e. $(a+ib)(c+id) = (ac-bd) + i(bc+ad)$) and geometric (i.e. "multiply the lengths and add the angles") rules for complex multiplication are equivalent (pages 8-10). The proof goes like this:

*

*Geometric to symbolic: The symbolic rule is determined as soon as we say that $i^2=-1$ and that the distributive property holds, so it suffices to show that these two properties hold. The former is true since $i$ has length $1$ and angle $\pi/2$, so squaring it produces length $1$ and angle $\pi$, i.e. $-1$. The latter is true since rotations and dilations preserve parallelograms.

*Symbolic to geometric: $iz = i(x+iy) = -y + ix$ according to the symbolic rule, so multiplication by $i$ is equivalent to a right-angle rotation. Using the distributive property we then have $(x'+y'i)z = x'z + y'(iz) = x'z + y'(z \text{ rotated by right angle})$. Drawing the picture, the triangle formed by points $(0,0), (x',0), (x',y')$ is now similar to the triangle formed by points $(0,0), x'z, (x'+y'i)z$, which shows that the length has been multiplied by the length of $x'+y'i$ and that the angle has been rotated by the angle of $x'+y'i$.

My basic question is something like: In what sense was the above a proof? How can we formalize what must be shown, and how does the above prove that?
One basic confusion I have is that the structure of the theorem at a high level looks like $A=B$  (symbolic=geometric), and yet we are having to show the equivalence in both directions. If we want to show $A=B$, we only need to show $A=B$ and not $B=A$ also. Maybe another way to state this is that it seems we just need to define complex multiplication in one way, then prove the other rule as a theorem, which seems to require only one direction. So in one sense by showing two directions the above proof seems overkill.
Okay so what are we trying to show? It seems like we have two representations of complex numbers, the algebraic/symbolic one $\mathbf C$ and the geometric one $\mathbf C'$. And we have multiplication $\cdot$ defined on $\mathbf C$, and multiplication $\cdot'$ defined on $\mathbf C'$. We want to show that there is some bijection $f : \mathbf C \to \mathbf C'$ such that $f(z\cdot w) = f(z)\cdot' f(w)$ for all $z,w\in \mathbf C$.
How does the above proof produce such an $f$? The "Geometric to symbolic" direction first shows that $f(i)^2 = f(-1)$. It then shows that $f(z)(f(w) + f(u)) = f(z)f(w) + f(z)f(u)$. If we assume $f$ interacts well with addition and multiplication by real numbers, we have $f(a+ib)f(c+id) = f((ac-bd) + i(ad+bc))$. But are we allowed to assume this? If so it sounds like we are done, and if not then I'm not sure how this direction of the proof is useful. For "Symbolic to geometric" direction I'm not sure how to translate it in terms of $f$.
 A: We have two models for the complex numbers:

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*In the geometric model, every complex number is a position vector $\vec{v}$ in the plane.

*In the symbolic model, every complex number is an expression of the form $a + i b$ for real numbers $a$, $b$.

The additive operations in the two models are defined as follows:

*

*In the geometric model, the sum of two complex numbers, $\vec{v}$ and $\vec{w}$, is their vector sum, $\vec{v} + \vec{w}$, which is defined by the parallelogram law.

*In the symbolic model, the sum of two complex numbers, $a+ib$ and $c+id$, is defined to be $(a+c)+i(b+d)$.

The multiplicative operations in the two models are defined as follows:

*

*In the geometric model, the product of two complex numbers, $\vec{v}$ and $\vec{w}$, is the vector $\vec{z}$ with $\lvert \vec{z} \rvert = \lvert \vec{v} \rvert \cdot \lvert \vec{w} \rvert$ and $\angle \vec{z} = \angle \vec{v} + \angle \vec{w}$.

*In the symbolic model, the product of two complex numbers, $a+ib$ and $c+id$, is defined to be $(ac-bd)+i(ad+bc)$.


Now, what Needham actually shows in the first part is the following.
Consider the bijection $f \colon \mathbb{C}_{\mathsf{GEOM}} \to \{ a + ib : a, b \in \mathbb{R} \}$ given by $f(\vec{v}) = v_x + i v_y$, where $\vec{v}$ is the position vector of the point $(v_x, v_y)$, and the codomain is just the set of all expressions of the form $a+ib$ for $a, b \in \mathbb{R}$. Then, we can transport the structure of $\mathbb{C}_{\mathsf{GEOM}}$ onto the codomain by defining $(a+ib)+(c+id) = f(f^{-1}(a+ib)+f^{-1}(c+id))$ and $(a+ib) * (c+id) = f(f^{-1}(a+ib) * f^{-1}(c+id))$.
Then, Needham shows that the resultant structure on the codomain is precisely $\mathbb{C}_{\mathsf{SYMB}}$.

Similarly, in the second part Needham shows the following.
Consider the bijection $g \colon \mathbb{C}_{\mathsf{SYMB}} \to \{ \vec{v} = (v_x, v_y) : v_x, v_y \in \mathbb{R} \}$ given by $g(a + ib) = \vec{v}$, where $\vec{v}$ is the position vector of the point $(a, b)$, and the codomain is just the set of all position vectors in the plane. Then, we can transport the structure of $\mathbb{C}_{\mathsf{SYMB}}$ onto the codomain by defining $\vec{v} + \vec{w} = g(g^{-1}(\vec{v})+g^{-1}(\vec{w}))$ and $\vec{v} * \vec{w} = g(g^{-1}(\vec{v}) * g^{-1}(\vec{w}))$.
Then, Needham shows that the resultant structure on the codomain is precisely $\mathbb{C}_{\mathsf{GEOM}}$.

Consider the first part. (Let $\vec{\iota}$ be the position vector of the point $(0,1)$.) How does Needham show that the resultant structure is indeed $\mathbb{C}_{\mathsf{SYMB}}$?

*

*Note that $f(\vec{\iota}) = i$ by the definition of $f$.

*Needham notes that it suffices to check that $i^2 = -1$ and that the distributive property holds.

*Since we have just transported the structure, $i^2 = -1 \iff \vec{\iota}^2 = \vec{-1}$, where $\vec{-1}$ is the position vector of the point $(-1,0)$.

*Similarly, the distributive property holds for the codomain with the transported structure iff it holds for the domain $\mathbb{C}_{\mathsf{GEOM}}$.

*Needham already showed how $\vec{\iota}^2 = \vec{-1}$ in an earlier section, and he shows that the distributive property holds in $\mathbb{C}_{\mathsf{GEOM}}$ to complete the proof.

*Of course, one also needs to verify that the additive structure matches, which Needham leaves as an exercise. By the transport of structure, we have by definition that $(a+ib)+(c+id) = f(f^{-1}(a+ib)+f^{-1}(c+id))$, that is, $f(\vec{v}) + f(\vec{w}) = f(\vec{v}+\vec{w})$ if $f(\vec{v}) = a+ib$ and $f(\vec{w}) = c+id$, so one needs to check that $f(\vec{v}+\vec{w}) = (a + c) + i(b + d)$. And, this is true by the way the parallelogram law of vector addition is defined.


For the second part, Needham shows that the resultant structure on the codomain is $\mathbb{C}_{\mathsf{GEOM}}$ as follows (again, let $\vec{\iota}$ be the position vector of the point $(0,1)$.):

*

*First, the verification that the additive structure matches is left as an exercise. (We have by definition that $\vec{v} + \vec{w} = g(g^{-1}(\vec{v})+g^{-1}(\vec{w}))$, and we can again complete the verification using the parallelogram law of addition.)

*For the verification of the multiplicative structure, it suffices to check that for a fixed element $\vec{v}$, the map $\vec{w} \mapsto \vec{v} * \vec{w}$ rotates $\vec{w}$ by $\angle \vec{v}$ in the counter-clockwise direction and expands it by $\lvert \vec{v} \rvert$.

*Needham first verifies this when $\vec{v} = \vec{\iota}$, and then uses the distributive property of $\mathbb{C}_{\mathsf{SYMB}}$ to show this for general $\vec{v}$.


Lastly, I believe the point that Needham is trying to emphasise here by transporting the structures from one model to the other and showing the equivalences (note that $g \circ f$ is the identity map on $\mathbb{C}_{\mathsf{GEOM}}$ and $f \circ g$ is the identity map on $\mathbb{C}_{\mathsf{SYMB}}$) is that neither is more fundamental than the other and that one should keep both models in mind when studying complex analysis since one viewpoint can be more useful than the other depending on the problem at hand. In fact, the entire book is based on the idea that the geometric viewpoint is far more useful than it is given credit for.
