Why can complex numbers be written in exponential form? $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.
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?
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$\begingroup$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.
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 $\begingroup$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 $f(t)=e^{it}$.
$\endgroup$ 17 $\begingroup$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:
- It's easy to use that form to read off the length and angle of your complex number
- 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.
- 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
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.
$\endgroup$ $\begingroup$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 $t$.
$\endgroup$ $\begingroup$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 $x \in \mathbb{C}$.
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