What's the magnitude of a real number?

Question

As a student of mathematics (first year master degree) I have to admit that I'm somewhat ashamed to ask this.

We know that if $z=x+iy$ is a complex number then we can identify it as $z=r\cdot\exp(i\theta)$. But what if $z$ is real - in other words its $y$ equals 0? Then $z=r\exp(i\cdot0)=r$ and this means that $z$ would be equal to its magnitude $r$ if $z$ is positive. But what if $z$ is negative? We know that the magnitude is always positive and so we'll get $z$(negative) = $r$(positive)?

I'm sure there's something I'm missing here.

Answer 1

If $z$ is real, then $\theta =0$ when $z > 0$ and $\theta = \pi$ when $z < 0$.

So $e^{i\theta} = 1$ when $z > 0$ and $e^{i\theta} = -1$ when $z < 0$.

Answer 2

In both cases, the magnitude of a number is simply its distance from zero. If $z=x+iy$, this is simply $\sqrt{x^2+y^2}$, by Pythagoras. So for a real, $$ \lvert x \rvert = \lvert x+0i \rvert = \sqrt{x^2} \geq 0. $$

Answer 3

It might be helpful to plot your complex number in the complex plane. Remember theta is found by considering a line segment from the origin to your complex number and finding the angle measure in standard position. So plotting a negative real number in the complex plane would give you theta=$\pi$.

Answer 4

In the definition $z = r e^{j\theta}$, by definition, the magnitude $r$ $\ge 0$. If you have a $z$, such that $\Im(z) = 0$ and $\Re(z) < 0$, then you must have $z = |z|e^{j\pi}$. The angle $\theta = \pi$ points in the direction of the negative real axis.

Answer 5

By Euler's identity, $$e^{i\pi}=-1,$$

which is how the sign gets corrected, via $\theta=\pi$.

Answer 6

As others explained, for a real negative number $z$ you have $\theta=\pi$ and $r=|z|=-z$, giving $z=(-z)\exp(i\pi)=(-z)\cdot(-1)=z$, and it all adds up.

But if I may, I would like to address your opening sentence:

As a student of mathematics (first year master degree) I have to admit that I'm somewhat ashamed to ask this.

First, a joke.

Dad, why is the sky blue?

I don't know, son.

Dad, how do airplanes fly?

I don't know.

Dad, what is in the center of the earth?

I don't know.

Dad, is it ok that I ask so many questions?

Of course! If you don't ask, how will you ever learn anything?

It's generally a good idea to ask questions when you're curious or confused about something.

And though I'd expect a student of your level to find a question like this trivial, we all have gaps in knowledge or mental blockages from time to time.

But I have to say - with the unintended risk of making you feel worse than you might already do - that what I find most troublesome here is that you were not able to reconcile this confusion on your own.

A mathematician should know, from experience, how to tackle problems. For a question like this, you might have wanted to try out taking some arbitrary complex numbers and figuring out their arguments. You would see that complex numbers which are close to the negative real number line have arguments close to $\pi$, and it should have dawned on you that the argument for negative real numbers must be $\pi$ rather than $0$ (or $-\pi$, depending on your choice of branch).

Not sure what you can do about it, since as mentioned this is something that comes from experience (which you should already have by now...). But perhaps focused study on problem solving techniques could help. I heard good things about George Pólya's "How to Solve It", though I haven't read it myself.

Answer 7

Since it has not been pointed out so far, it is crucial to realize that while cartesian coordinates are unique, polar coordinates are not unique, because the pair $(r\cos(t),r\sin(t))$ is exactly the same as $(r\cos(t+2πn),r\sin(t+2πn))$ for any integer $n$. Furthermore, $(0\cos(t),0\sin(t)) = (0,0)$ for any real $t$. So if we want to represent points by polar coordinates $(r,t)$, we usually stipulate a range for $t$. Commonly $0 \le t < 2π$ or alternatively $-π < t \le π$. In either case, if we require $r>0$ then $(r,t)$ is unique for any point except the origin, because the distance from origin and the angle of the ray from the origin are uniquely determined. There is still no unique polar coordinates for the origin.

One could choose to allow $r$ to be negative and restrict $t$ further to $0 \le t < π$, so that the point $(-1,0)$ (cartesian coordinates) can have 'polar' coordinates $(-1,0)$ rather than $(1,π)$. However, this is not as elegant or convenient as the usual definition of polar coordinates.