The Real number line is in one dimension. If you want to map a complex number, you would have to add a second dimension to that number line- the Imaginary-axis.
The Imaginary-axis is always perpendicular to the Real-axis. Here is my question:
Would you still be able to use the complex plane if the imaginary-axis wasn't perpendicular to the Real-axis? In other words, is it still possible to prove theorems involving complex numbers (in a geometrical way) if the Imaginary-axis wasn't perpendicular to the Real-axis?
For example: Could you prove Euler's formula if the Imaginary-axis was tilted to a 45 degree angle?
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$\begingroup$Sure, you could construct the complex numbers with some other "basis" besides $1$ and $i=\sqrt{-1}$. What about $1$ and $k={-1}^{1/4}= \frac{1+i}{\sqrt{2}}.$ Addition will still work and you will have some other rule of multiplication:
For example, with $z_1= a_1+ k b_1$ and $z_2=a_2 + k b_2$,
$$\begin{aligned} z_1 z_2 & =(a_1 + k b_1)(a_2 + k b_2) \\& = a_1a_2+k (a_1 b_2+ a_2 b_1)+ k^2 b_1 b_2 \\& = a_1 a_2 + k (a_1b_2 + a_2b_1)+(\sqrt{2}k-1)b_1b_2\\&= a_1a_2-b_1b_2+k(a_1b_2+a_2b_1+\sqrt{2}b_1b_2)\end{aligned} $$
So this other "complex number system" is closed under addition and multiplication.
Similarly, we can show that the "polar form" is okay too.
$\endgroup$ $\begingroup$In the complex plane, the multiplication by i can be represented as a rotation by 90º counterclockwise, and, therefore, if you want multiplication to have that meaning, it is required that i·i=-1 and therefore the imaginary axis intersects the real axix with an angle of 180º/2=90º.
If the axes are not perpendicular, you can not satisfy the equality a_α·b_β = (a·b)(α + β), because i·i = 1_φ·1_φ!=1_90º·1_90º=1_180º=-1
However, you can still represent addition and substraction if your axes are not orthogonal.
$\endgroup$ 2 $\begingroup$Short answer: it's not necessary, just uniquely insightful.
The compound angle formulae of real trigonometry prove$$\operatorname{cis}x:=\cos x+i\sin x\implies\forall x,\,y\in\Bbb R(\operatorname{cis}x\operatorname{cis}y=\operatorname{cis}(x+y)).$$We can extend this to $x,\,y\in\Bbb C$ if we define all functions by their Maclaurin series. So $\operatorname{cis}x$ is the unique solution of $f^\prime = if,\,f(0)=1$, i.e. $\operatorname{cis}x=\exp ix$.
None of this analysis cares what angle we put between the real and imaginary axes in a diagram (the diagram is only an invention for convenience), and as long as these axes aren't parallel each point in the plane has a unique real and imaginary part, and represents a unique complex number.
But since $\exp\frac{i\pi}{2}=i$, perpendicular axes have the unique benefit of allowing us to interpret the argument of $\operatorname{cis}$ as a polar angle. This has other benefits, such as identifying the roots of unity with a regular polygon's vertices, and making obvious the isomorphism between multiplying unit complex numbers and rotating in $2$ dimensions.
$\endgroup$ $\begingroup$It depends on how structured you make your "complex" numbers. As a pure algebraic field, the answer is "no" - but then again, as such, they also have no topology, either, and you cannot talk about things such as limits, convergence, or even relative size of one number versus another.
If, however, you expand the complex field to a valued field in the standard way, i.e. you include the absolute value function
$$|\cdot| : \mathbb{C} \rightarrow \mathbb{R}^{\ge0},\ \ \ |z| := \sqrt{z \bar{z}}$$
then this function encourages a natural metric structure, topological structure homeomorphic to the plane ($\mathbb{R}^2$), and then these in turn imply a natural notion of angle, and hence the imaginary and real axes are, indeed, perpendicular. This follows from that $|\cdot|$ is equivalent to the metric of a Riemannian manifold, namely the Euclidean plane, and hence by a result known as the Myers-Steenrod theorem, determines also its metric tensor $g$, and thus the angle concept. See:
So I'd say, "yes, the axes are perpendicular, to about the same extent that the real number '2' is at a distance of 1 unit from '1', and any similar such example". $|\cdot|$ is to $\mathbb{C}$ what the ordering relation $\le$ is to $\mathbb{R}$, more or less, in that it's what gives it its topological and metrical structure. You could try redefining that, of course, but you can also do the same for the reals as well, and so change both systems' geometric properties.
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