15
$\begingroup$

Why can complex numbers be written in exponential form? $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.

I have studied that the exponential form of a complex number $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.

Can someone explain why?

$\endgroup$
6
  • $\begingroup$ @A---B: I have reverted the edit which changed the title of the question to “how to change from polar to exponential form…” — it does not seem to be what the question is asking, so changes the meaning of the question. $\endgroup$ Commented Sep 19, 2017 at 19:32
  • $\begingroup$ @ShreevatsaR Check now. $\endgroup$
    – user312097
    Commented Sep 19, 2017 at 19:33
  • 3
    $\begingroup$ @A---B I am not sure “Why complex numbers are written in expoenential form?” is fine either (besides introducing a typo), because the question asks “I have studied that … can be written… can somebody explain why?”, so it may be asking about why complex numbers can be written in exponential form, rather than why they are. $\endgroup$ Commented Sep 19, 2017 at 19:37
  • $\begingroup$ @ShreevatsaR Yes I think you are correct. $\endgroup$
    – user312097
    Commented Sep 19, 2017 at 19:44
  • $\begingroup$ There is deeper justification for the equivalence of polar form and exponential form of a complex number (and this is beyond algebra). For an introduction to complex numbers this equivalence can be thought of a mnemonic to help remember rule for multiplication of complex numbers: multiply moduli and add argument. $\endgroup$
    – Paramanand Singh
    Commented Sep 20, 2017 at 4:28

6 Answers 6

16
$\begingroup$

Here's a rather elegant proof.

The function $f : t\mapsto \cos t+i\sin t$ is differentiable and satisfies \begin{align*} f'(t) &= i\,f(t) \\ f(0) &= 1 \end{align*} Now let's solve it.

We have $f(0) = 1$ and

$$f'(t) = (\cos t+i\sin t)' = -\sin t+ i\cos t = i(\cos t+i\sin t) =if(t) $$ Now let us solve this differential equation $$f'(t) = if(t)\Longleftrightarrow e^{-it}f'(t) -ie^{-it} f(t)=0 \Longleftrightarrow \frac{d}{dt}\left(e^{-it} f(t)\right) = 0$$

That is $$e^{-it} f(t) = c\Longleftrightarrow f(t) = ce^{it}$$

But $f(0)=1 $ i.e $c=1$. Hence $f(t)=e^{it}$.

$\endgroup$
18
  • 4
    $\begingroup$ It's not clear what you mean by "$\ln(f(t))$". Someone wondering why $e^{it} = \cos t + i \sin t$ probably doesn't know what complex logarithm is; even worse, they may not even know that they don't know what it is! (And thus miss the fact that there's something nontrivial to say.) $\endgroup$ Commented Sep 20, 2017 at 11:38
  • 6
    $\begingroup$ I don't understand what you are trying to prove here, and what is your definition of $e^{it}$ while you do it. $\endgroup$ Commented Sep 20, 2017 at 12:45
  • 1
    $\begingroup$ It is a little easier to show that $t \mapsto e^{-it} f(t)$ is a constant. Then the proof becomes a one liner. $\endgroup$
    – copper.hat
    Commented Sep 20, 2017 at 14:07
  • 2
    $\begingroup$ When you use the power series definition of $e^z$ the result in question follows trivially from Taylor series for sine and cosine. So that part is easy. On the other hand how do you propose to analytically extend the definition of $e^{x} $? One of the ways of doing this is via power series. Other is to define $e^{x+iy} =e^{x} (\cos y+i\sin y) $ which again makes the question trivial. From your solution it appears that you assume that $g(t)=e^{it} $ satisfies the differential equation (in the beginning of your post). This makes it trivial as one can show that the equation has unique solution. $\endgroup$
    – Paramanand Singh
    Commented Sep 21, 2017 at 5:25
  • 4
    $\begingroup$ Continuing Paramanand Singh's comment: if we don't know what $e^{it}$ is, how do we know what its derivative is? That is, how do we know that it satisfies $\frac{\mathrm{d}}{\mathrm{d}t}f(t)=i\,f(t)$? $\endgroup$
    – robjohn
    Commented Sep 21, 2017 at 5:43
15
$\begingroup$

Lets consider a function from $\mathbb R\to \mathbb C$

$z(\theta) = \cos \theta + i\sin \theta\\ z(\theta)z(\phi) = (\cos \theta + i\sin \theta)(\cos \phi + i\sin \phi) = \cos(\theta + \phi) + i\sin (\theta+\phi) = z(\theta + \phi)$

That is a property of an exponential function. We do not know the base.

For some base:

$\exp (iy) = z(y) =\cos y + i\sin y$

and:

$\exp (x + iy) = \exp(x)\exp(iy) =\exp(x) (\cos y + i\sin y)$

And then you can define $e$ to be the required base. In much of complex analysis, it does not matter that it is the same $e$ as you have learned to be Euler's constant.

However, if you have taken calculus, you should recognize these Taylor expansions.

$e^x = \sum_\limits{n=0}^{\infty} \frac {x^n}{n!}\\ \cos x = \sum_\limits{n=0}^{\infty} \frac {(-1)^nx^{2n}}{(2n)!}\\ \sin x = \sum_\limits{n=0}^{\infty} \frac {(-1)^nx^{2n+1}}{(2n+1)!}$

what is

$e^{ix}$ ?

$e^{ix} = \sum_\limits{n=0}^{\infty} \frac {{ix}^n}{n!}\\ 1 + ix + \frac {(ix)^2}{2} + \frac {(ix)^3}{3!}+ \frac {(ix)^4}{4!} \cdots\\ 1 + ix + \frac {-x^2}{2} + \frac {-ix^3}{3!} + \frac {x^4}{4!} \cdots$

collect the real terms and the imaginary terms

$(1 - \frac {x^2}{2} + \frac {x^4}{4!}\cdots )+ i( x - \frac {x^3}{3!} + \frac {x^5}{5!} \cdots)\\ e^{ix} = \cos x + i\sin x$

Without calculus.

we can define $e = \lim_\limits{n\to\infty}(1+\frac {1}{n})^n\\ e^x =\lim_\limits{n\to\infty} (1+\frac {1}{n})^{nx} $

Make a substitution $m = nx$

$e^x =\lim_\limits{m\to\infty} (1+\frac {x}{m})^m $

Then look at what happens as $m = 1, 2,3, etc.$

We have already shown that multiplication of complex numbers multiplies the lengths and adds the angles.

enter image description here

As $m$ increases hopefully you can see how that sequence of line segments begins to lie on the curve of the circle.

and when $m$ is very large comes to rest on $\cos x + i\sin x$

$\endgroup$
0
9
$\begingroup$

There are several reasons. Even without going into the technical details of why it's correct, here is a small list of reasons for why might be a good idea:

  1. It's easy to use that form to read off the length and angle of your complex number
  2. It's easy to recognize the form and see that it is indeed meant to convey length and angle as opposed to, for instance, the width and height we see in the $a+bi$ form.
  3. The rules for complex multiplication means that hijacking the exponential notation and (mis)use the intuition you have from real exponentiation gives the correct results
$\endgroup$
6
$\begingroup$

In this answer, it is shown that $$ \lim_{n\to\infty}\left(1+\frac{i\theta}n\right)^n=\cos(\theta)+i\sin(\theta)\tag1 $$ Therefore, we can say $$ e^{i\theta}=\cos(\theta)+i\sin(\theta)\tag2 $$ We can also use the power series for $e^x$, $\cos(x)$, and $\sin(x)$ to derive $(2)$.

In any case, once we have $(2)$, any point on the unit circle in $\mathbb{C}$ can be represented as $e^{i\theta}$ for some $\theta\in\mathbb{R}/2\pi\mathbb{Z}$; $\theta$ is the argument of that point.

Furthermore, using $(2)$, we can write any point in $\mathbb{C}$ as $$ \begin{align} x+iy &=\overbrace{r\cos(\theta)}^x+i\,\overbrace{r\sin(\theta)}^y\\ &=re^{i\theta}\tag3 \end{align} $$ One important identity is $$ \begin{align} e^{i\theta}e^{i\phi} &=(\cos(\theta)+i\sin(\theta))(\cos(\phi)+i\sin(\phi))\\ &=(\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi))+i(\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi))\\ &=\cos(\theta+\phi)+i\sin(\theta+\phi)\\ &=e^{i(\theta+\phi)}\tag4 \end{align} $$ Equation $(4)$ tells us that we can combine imaginary exponents like we do real ones.

$\endgroup$
3
$\begingroup$

If $z=0$ it is clear that we can take $r=0$ and any value for $\theta$.

Note that any number on the complex unit circle can be written as $\cos t + i \sin t$ for some $t$.

Note that any non zero complex number $z$ can be written as $z= |z| {z \over |z|}$ and ${z \over |z|}$ lies on the complex unit circle.

If we let $r=|z|$, then we see that there is some $t$ such that $z = r (\cos t + i \sin t)$.

As to why $e^{it} = \cos t + i \sin t$, let $\phi(t) = e^{it} - ( \cos t + i \sin t ) $, note that $\phi(0) = 0$ and $\phi'(t) = i\phi(t)$. Hence $e^{-it} \phi(t)$ is a constant from which it follows that $\phi(t) = 0$ for all $t$.

$\endgroup$
0
$\begingroup$

It's a definition.

As to why it's a good definition, the answer comes from the fact that the Taylor series for exp(x) stays the same when we let $x \in \mathbb{C}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .