When I see examples of complex numbers on the complex plane (imaginary $y$-axis, real $x$-axis), we have some number $z = a + bi$ and almost every diagram, tutorial, says this maps to the point $a$ units over on the real line and then $b$ units up/down on the imaginary line which is labeled $..., -2i, -1i, 0, 1i, 2i, ...$.

Why is this so? It's treated like some obvious statement. But I don't see why it should be the case. It's like $z = a + bi$ is really being treated as a coordinate pair $(a, b)$.

  • $\begingroup$ Why should we use a regular number line to represent real numbers? It's mostly just a matter of visualization. $\endgroup$ – Sean Henderson Feb 21 '18 at 21:18

We use the complex plane because it provides a very helpful visualization of complex numbers. The reason we can use it is because each complex number $z = a + bi$ is completely determined by the real numbers $a$ and $b$. In a way, thinking about a complex number as a point in $\mathbb{R}^2$ is just a different way of notating that number. There's no danger of confusion or ambiguity, because everyone can agree that a given point $(a,b)$ corresponds to the complex number $a + bi$, and vice versa. This is explained also in Atmos's answer.

I think it's worth mentioning that there is actually more than just this correspondence between the elements of each set, however.

We can also think about different operations that can be done with complex numbers and vectors in $\mathbb{R}^2$, and we can compare these. We know that complex numbers can be added together, as can vectors in $\mathbb{R}^2$. Complex numbers and vectors in $\mathbb{R}^2$ both add in the same way. Compare the following, for $a,b, c, d \in \mathbb{R}$

$$ (a + bi) + (c + di) = (a + c) + (b + d)i \\ (a,b) + (c,d) = (a + c, b + d) $$

You can see that to add two complex numbers you can just think of each complex number as a point in $\mathbb{R}^2$ and add these points as you normally would. The result is the point in $\mathbb{R}^2$ corresponding to the sum of the complex numbers.

We can also scale complex numbers (multiply them by a real constant), and also scale vectors in $\mathbb{R}^2$. Again, this scaling behaves in the same way in $\mathbb{C}$ as it does in $\mathbb{R}^2$. Let $a,b,k \in \mathbb{R}$ and compare again

$$ k(a + bi) = ka + (kb)i \\ k(a,b) = (ka, kb) $$

We see that to scale a complex number by a real constant, we can just think of the complex number as a point in $\mathbb{R}^2$, scale that point as we normally would, and the result is the point in $\mathbb{R}^2$ corresponding to the scaled complex number. If you'd like, the precise way of saying all of this is to say that $\mathbb{C}$ and $\mathbb{R}^2$ are isomorphic as $\mathbb{R}$-vector spaces. So you can be confident that the things you do with vectors (i.e. scalar multiplication, and vector addition) are the same in $\mathbb{C}$ and $\mathbb{R}^2$, so we are permitted to use this helpful visualization.

Edit: Here's an explanation of some extra behavior exhibited by $\mathbb{C}$ that is not compatible with the behavior of vectors in $\mathbb{R}^2$.

Above I did not explain the whole story of the multiplication operation. I explained how scaling by a real number works, but I did not mention what happens when you scale by an arbitrary complex number. Multiplication of complex numbers has a nice geometric interpretation when thinking of the elements of $\mathbb{C}$ as points on the plane. Each complex number $z = a + bi$ has associated to it a magnitude and an angle as measured counterclockwise from the positive $x$ axis. When multiplying complex numbers $z$ and $w$, the result is a new complex number with magnitude $|z|\cdot|w|$ and an angle that is the sum of the angles of $z$ and $w$. That is, multiplication of complex numbers corresponds to scaling and rotating on the complex plane. I explained a special case of this above when I talked about scaling by a real number. A real number has an angle of zero (it is on the $x$ axis), so multiplying by a real number only scales and does no rotation.

I mentioned that this operation is not compatible with a corresponding operation on $\mathbb{R}^2$. On one hand, this is the case because we are thinking of $\mathbb{R}^2$ as a vector space. And in a vector space, multiplication of two vectors need not be defined. So, this operation of multiplication is some sort of extra property possessed by the complex numbers that is not possessed by the vector space $\mathbb{R}^2$. This doesn't really change anything that I've already said, however. We can still think of complex numbers as points in $\mathbb{R}^2$. We just have to remember this extra rule for what happens to two points when we multiply them together.

And finally, some more detail if you still happen to be interested. We can define multiplication of elements in $\mathbb{R}^2$, and the natural way to do this is to say that for $a,b,c,d \in \mathbb{R}$

$$ (a,b) \cdot (c,d) = (ac, bc) $$

This makes $\mathbb{R}^2$ into a ring. This way of multiplying elements, however, is not the same as the way we multiply elements in $\mathbb{C}$ because in $\mathbb{C}$ we have the relation that $i^2 = -1$, which changes things slightly. The precise way to state this difference is to say that $\mathbb{R}^2$ and $\mathbb{C}$ are not isomorphic as rings. Again though, this often doesn't matter to us, because in the situations where we want to think of $\mathbb{C}$ as $\mathbb{R}^2$ we are thinking of their compatibility as vector spaces, not as rings.

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  • $\begingroup$ This is really good, plus one. But you should have mentioned the geometric significance and interpretation of multiplying by a complex number, not just a real number. $\endgroup$ – Lubin Feb 21 '18 at 18:17
  • $\begingroup$ @Lubin You are right, I should add that in. I’ll update it when I’m back at my computer. $\endgroup$ – wgrenard Feb 21 '18 at 18:20
  • $\begingroup$ I still don't understand why we say $(a,b)$ in $\mathbb{R}^2$ is the same as $a + bi$ in... whatever the complex plane is. $\mathbb{I} \times \mathbb{R}$? $\endgroup$ – user525966 Feb 21 '18 at 18:22
  • $\begingroup$ @user525966 You would say $a + bi \in \mathbb{C}$ The complex plane $\textit{is}$ $\mathbb{R}^2$ but with the labels changed. Usually you label the axes "real" and "imaginary" instead of $x$ and $y$, but the labels are inconsequential. In your question you mention "It's like $z=a+bi$ is really being treated as a coordinate pair $(a,b)$." This is correct. We freely go back and forth between the notations because we can all agree that the two notations represent the exact same information without any ambiguity. $\endgroup$ – wgrenard Feb 21 '18 at 18:50
  • $\begingroup$ Maybe I am confusing myself here, the usual Cartesian plane, where $x$ is a real axis, $y$ is a real axis, this is $\mathbb{R}^2$, but the complex plane, where the horizontal axis represents real numbers and the vertical represents imaginary numbers, is also $\mathbb{R}^2$? How do we distinguish between the two? $\endgroup$ – user525966 Feb 21 '18 at 18:55

Because to every point of the plane $(x,y)$ corresponds a unique complex number $z=x+iy$. We state that there exists a bijection between the set of complex number $\mathbb{C}$ and the real plane $\mathbb{R}^2$, that's why we often call it the complex plane.

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  • $\begingroup$ Are those $x$ and $y$ terms meant to be different? Or are you saying that something like $(3, 4)$ corresponds to $z = 3 + 4i$? $\endgroup$ – user525966 Feb 21 '18 at 17:18
  • $\begingroup$ $(3,4)$ corresponds to $3+4i$. That's why the axis $x$ defined by all points of the form $(a,0)$, $a \in \mathbb{R}$ is, infact, the real number and those of the form $(0,a)$, which are the $y$ axis, is the pure imaginary numbers. $\endgroup$ – Atmos Feb 21 '18 at 17:20
  • $\begingroup$ Now I'm lost, why would we map the real $y$ axis in the real plane to the imaginary axis in the complex plane? $\endgroup$ – user525966 Feb 21 '18 at 17:28
  • $\begingroup$ You are confusing the representation of a complex number with your interpretation of the "plane". Think of the x and y axes instead as distances from the origin, and nothing else. Then, the number $3i$ is just a point represented as moving 3 units along $+y$. This distance is a real number. Now interpret that point as a complex number - considering it as a coordinate just tells you how far to move in the complex plane, but you're not done until you put additional interpretations on what that coordinate means. $\endgroup$ – Michael Stachowsky Feb 25 '18 at 12:58

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