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Definition $1$: Suppose $A$ and $B$ are two complex numbers. $A+B$ is the fourth vertex of a parallelogram with vertices $O, A, B$, where $O$ is the origin.

Definition $2$: Suppose $A$ and $B$ are two complex numbers written as $A = a_0 + a_1i$ and $B = b_0 + b_1i$. Then $A+B = (a_0 + b_0) + i (a_1 + b_1)$.

Def $1$ $\implies$ Def. $2$

Let $A = a_0 + a_1i$ and $B = b_0 + b_1i$ and let $P = A+B$. By definition, $0, A, B, P$ is a parallelogram, so side $PB$ must be parallel to and the same length as side $AO$. If we add $b_0$ to $a_0$ and $b_1$ to $a_1$, we end up at a point $Q$, such that $QB$ is parallel to and has the same length as $AO$. Therefore $Q = P$.

I know I have to do the other direction as well, but I want to ask about this one first before I work on the other one. I had a little trouble coming up with it, because the concept is so simple that I am having trouble expressing it in more primitive terms. My concern is that the proof doesn't really prove or explain anything, it's just a convoluted way of saying "look at the picture, its obvious".

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    Yes, it is correct. Once one shows the isomorphism $\;\Bbb C\cong\Bbb R^2\;$ as vector spaces, say (or as additive groups), and one knows how vectors are added in $\;\Bbb R^2\;$ , there's nothing really to prove. – 2017-02-26
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    @DonAntonio One of the problems is that I'm not really given a formal structure to work in. The book I am reading is Tristan Needham's Visual Complex Analysis. He just draws the complex numbers on a plane without giving any formalisms. – 2017-02-26
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    Ok, but he there begins with a rather short historical sketch and then on page two he makes clear the all powerful, important and basic identification $\;a+ib\to (a,b)\;$, between $\;\Bbb C\;$ and $\;\Bbb R^2\;$ . Well, then you work there as you did in analytic geometry with the real plane... – 2017-02-26
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    You could argue that $C=A+B$ by **Def 2** $\iff C-A = B-0 \iff AC$ and $OB$ are parallel and equal $\iff OABC$ is a parallelogram. Problem with the direct implication **1** $\implies$ **2** is that without complex addition (and presumably subtraction) defined, how you are expected to translate the parallelogram condition into a relation between complex numbers depends on prior context. – 2017-02-26
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    @dxiv I'm not sure I follow "without complex addition (and presumably subtraction) defined, ...". Aren't Definitions $1$ and $2$ definitions of complex addition? – 2017-02-26
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    @Ovi At the point where you start "**Def $1$ $\implies$ Def. $2$**" you don't have any definition for complex addition, yet. To prove the implication, you need to translate the parallelogram condition into something about complex numbers, but you must do so without using addition or subtraction. This eliminates the usual geometric interpretation which associates vector $\overrightarrow{AB}$ with the difference $B-A$ between complex affixes. – 2017-02-26

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