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I understand what the polar form of complex numbers means, but I don't understand why it's justified. Complex numbers, by definition, have nothing to do with a complex plane. The complex plane is simply a graphical way to represent them. So representing complex numbers in polar form seems to me to be a leap.

My only guess is that it's a leap that's useful and internally consistent, somewhat analogous to expressing negative exponents as exponents of reciprocals and reciprocal exponents as roots. Doing so retains the laws of exponents, so we define such exponents that way because it's useful and consistent.

The polar form of complex numbers just seems like a bigger leap, I guess. It's always prefaced by, "If we represent a complex number on the complex plane…", which seems arbitrary itself. I'm hoping that I'm missing something and that there's a more concrete justification.

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  • $\begingroup$ Euler's identity! $\endgroup$ – Parcly Taxel Sep 2 '17 at 5:32
  • $\begingroup$ Right, beautiful. Had a t-shirt with it on it in college. But, it depends on the polar form, as do all expressions involving powers to complex numbers, which is what reminded me of the question. $\endgroup$ – Chuck Sep 2 '17 at 5:34
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    $\begingroup$ Complex numbers, by definition, have nothing to do with a complex plane That's a bit too strong of a statement. $\,\mathbb{C}\,$ is isomorphic to $\,\mathbb{R}^2\,$, and $\,\mathbb{R}^2\,$ has a natural representation as a 2D plane. $\endgroup$ – dxiv Sep 2 '17 at 5:37
  • $\begingroup$ Did you have a problem with polar coordinates in calculus? There's nothing going on here beyond that. $\endgroup$ – zhw. Sep 2 '17 at 6:03
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    $\begingroup$ You can use the polar form of complex numbers without actually thinking geometrically at all. The identity $re^{i\theta}=r\cos\theta+ir\sin\theta$ can be proven purely algebraically. (Alternatively, you can forego working with complex exponentials entirely and still use algebra only to show everything you need, e.g. $(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)=\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)$ follows just from the sine and cosine of sum formulas). $\endgroup$ – Carmeister Sep 2 '17 at 6:24
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It's possible to show that we can work with the polar form of complex numbers without actually using any geometry at all!

$\DeclareMathOperator{\cis}{cis}$ First, given a non-zero complex number $z=a+bi$, we can let $r=\sqrt{a^2+b^2}$ and rewrite $z$ as $$z=r\left(\frac ar+\frac bri\right).$$ From the definition of $r$, we know that $(a/r)^2+(b/r)^2=1$. Now we want to show that for any real numbers $x$, $y$ with $x^2+y^2=1$, there is a $\theta$ such that $\cos\theta=x$ and $\sin\theta=y$. The easiest way to make this argument may be geometrically, but we can still do so without using geometry, by defining $\theta$ by $$\theta=\begin{cases}\arccos x&y>0\\ -\arccos x&y<0\\ 0&y=0,\,x=1\\ \pi &y=0,\,x=-1\end{cases}$$ (where $\arccos$ is assumed to take values in $[0,\pi)$). It's not all that pretty, but hopefully you can convince yourself that it works.

Going back to our complex number $z$, if we apply our lemma to $x=a/r$ and $y=b/r$, we can find a $\theta$ with $\cos\theta=a/r$ and $\sin\theta=b/r$, so $z=r(\cos\theta+i\sin\theta)$. So we have showed the existence of the polar form of a complex number! The quantity $\cos\theta+i\sin\theta$ is sometimes abbreviated as $\cis\theta$; I'll use that in the rest of the answer.

Now that we have the existence of the polar form, can we show that it's useful? What happens when we multiply two complex numbers in polar form? Suppose $z_1=\cis\theta_1$ and $z_2=\cis\theta_2$. Then, $$\begin{align*}z_1z_2&=\cis\theta_1\cis\theta_2\\ &=(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)\\ &=(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\cos\theta_1\sin\theta_2+\cos\theta_2\sin\theta_1)\\&=\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\\&=\cis(\theta_1+\theta_2)\end{align*}$$ All we've used are the addition formulas for $\sin$ and $\cos$! A similar derivation shows that if we divide instead, we get $z_1/z_2=\cis(\theta_1-\theta_2)$.

Using polar form becomes especially useful if we want to take powers of complex numbers. Computing $(1+i)^5$ by multiplying everything out is a pain. But if we instead recognize that $1+i=\sqrt2\cis(\pi/4)$, then it's much easier to find $(1+i)^5=4\sqrt2\cis(5\pi/4)=-4-4i$.

Once you've proven Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$ you can dispense with the $\cis$ and use $re^{i\theta}$ instead of $r\cis\theta$, but the nice thing about using $\cis$ is that it can be done without any knowledge of calculus or exponentials. I hope that this sheds some light on the subject!

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Remember that real numbers, i.e., $\mathbb{R}$, can be thought of as being somewhere on the real line (or, alternatively phrased: each real number can be associated with a point on the real number line).

Similarly, $\mathbb{R}^2 = \mathbb{R}\times\mathbb{R}$ can be associated with the plane.

There is a natural bijection between $\mathbb{R}^2$ and $\mathbb{C}$ that sends $(a, b)$ to $a + bi$. In this way, one can think of the complex numbers (as a set) as consisting of two real numbers (i.e., the "real part" and the "imaginary part"). And so, just as you can associate $\mathbb{R}^2$ with the plane, so, too, can you do so for $\mathbb{C}$.

Having a geometric representation can, of course, help you solve some problems that involve complex numbers. As an example: Compute $\sqrt{i}$. How do you start such a problem?

Let us use your noticed preface:

If we represent a complex number on the complex plan...

...then $i = 0 + 1i$ would be located at $(0, 1)$. In polar coordinates, this is $(1, \pi/2)$. Taking the square root may now seem a bit easier, since we see that "multiplying" the point at $(1, \pi/4)$ by itself will retain the modulus of $1$ and rotate by $\pi/4$ again to arrive at the aforementioned $(1, \pi/2)$.

So $(1, \pi/4)$ serves as an answer. Converting back to the standard form for complex numbers, this is the point found at $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$. We can now compute the square of this number algebraically to verify that, indeed, it yields the desired value of $i$:

$$(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)^2 = \frac{2}{4} + 2\frac{2}{4}i - \frac{2}{4} = i$$

Of course, square roots come in pairs, so we'll want the negation of this number as well, and you can check that the number written as $-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$ would be located in the complex plane with polar coordinates $(1, 5\pi/4)$, so that squaring it ends up at $(1, 10\pi/4) = (1, 2\pi + \pi/2) = (1, \pi/2)$ so that we are, again, back at the polar coordinates for $i$ as desired.

This answer is by no means the end of the story; but, suffice to say that just as complex numbers can be associated with an ordered pair of real numbers and, therefore, associated with Cartesian coordinates, thus, represented in the plane as $(a,b)$, so, too, can they be represented in this (complex) plane using polar coordinates $(r, \theta)$ in a manner that can turn out to be useful for solving many problems (of which only a modestly simple example is given above).

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  • $\begingroup$ This answer helps, but doesn't quite justify it for me yet, although it does demonstrate that it works. That we can convert a complex number to polar form to solve problems, convert the answer back and it will work. So yes, it's consistent and useful. But I find it amazing that it works! It seems like magic to me. I don't usually find math magical. Fascinating, fun, but not magical. In every other instance, when I understand some part of math, I can look at it and say, "Yea, that makes sense." Not (yet) so with the polar form of complex numbers. $\endgroup$ – Chuck Sep 2 '17 at 6:15
  • $\begingroup$ @Chuck I share in your wonder! In fact, you may wish to check the top answer to MO 14574, which is a question entitled: "Your favorite surprising connections in Mathematics." $\endgroup$ – Benjamin Dickman Sep 2 '17 at 6:18
  • $\begingroup$ We can, if we want, define the set of complex numbers to be the complex plane (a.k.a. the Gauss-Argand plane). They are usually considered to be the same thing. $\endgroup$ – DanielWainfleet Sep 2 '17 at 17:37

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