I have always had this question in mind. I am aware that a complex number is of the form a+ib can be represented in a complex plane where the real part "a" is represented on the x-axis or the real axis and "b" is represented on the y-axis, the imaginary axis (I do not understand the concept of complex number very deeply; which is why its a matter of interest for me; so please correct me if this understanding of mine is wrong in any way), but what if a=0 and b=i? where would such a number be plotted on the complex plane?
Where does $i^2$ lie on the imaginary axis
complex numbers
Related Solutions
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.
In principle we could do what you suggest -- take $\mathbb R^2$ and associate every point $(x,y)$ to the number $x+13k$. Though the trouble with that particular plan is that each number now represents many different points -- for example, $(13,0)$ and $(0,1)$ and $(26,-1)$ are now all associated to the number $13$. This means that we can't use the scheme for anything where we calculate a number and that number points to exactly one point in the plane.
We could, however, do something more general. Take some field that extends $\mathbb R$, pick some element $\alpha$ in that field, and then represent $(x,y)\in\mathbb R^2$ by $x+\alpha y$.
As it went for $13$, if we pick $\alpha\in\mathbb R$, then we get something where a number doesn't represent a unique point. Suppose, however, that we steer clear of that case, and furthermore that we end up in the lucky situation that every element of the field represents some $(x,y)$ in the plane.
Something wonderful happens then -- namely, we can then prove (though not in the space left for me in this margin) that the field we're using must be isomorphic to $\mathbb C$ -- in other words the field is essentially the complex numbers, just called something different. In particular, somewhere in the plane there is an $(x,y)$ whose corresponding number behaves exactly like $i$.
So we could actually have said: Pick some complex number $\alpha$, and let $(x,y)$ correspond to $x+\alpha y$. As long as $\alpha$ is not real, this will give us a perfectly good one-to-one correspondence between points and complex numbers.
Now, among all the possible choices of $\alpha$ it turns on that exactly when $\alpha=i$ or $\alpha=-i$ we get the additional nice property that multiplication by any fixed nonzero complex number will correspond to a transformation of the plane that takes geometric figures to similar geometric figures.
Having multiplication correspond to similarity transforms is a pretty nifty property, which is a reason to prefer the representation $x+iy$ over other possible $x+\alpha y$.
Best Answer
$a+bi$ has coordinates $x = a, y = b$ when $a$ and $b$ are real. When they are not real, you will have to do multiplication of complex numbers to write the thing in standard form.
So $i^2 = -1 = -1 + 0i$ which has coordinates $(-1,0)$. It does not lie on the imaginary axis at all.