6
$\begingroup$

I want to define $e^{i z}$ without using power series. My approach was as follows:

Start with $$\lim_{n \to \infty} \left(1+\frac{i}{n}\right)^{nz}$$ Then make the substitution $$\frac{1}{m} = \frac{i}{n},\ \text{ so } n=mi.$$ And we have $$\lim_{n \to \infty} \left(1+\frac{1}{m}\right)^{miz}$$ This almost works, but for $n \to \infty$ we have $m \to i \infty.$ So at this point we can't say $$\left[\lim_{m \to \infty} \left(1+\frac{1}{m}\right)^{m} \right]^{iz}=e^{iz}.$$ Because we don't have $m \to \infty.$

So my next idea was to try $$\lim_{x \to 0} \left(1+ ix\right)^{z/x}$$ Then make the substitution $y=ix.$ Then for $x \to 0$ we have $y \to 0.$ So we can say $$\lim_{y \to 0} (1+y)^{iz/y}$$ It is tempting now to write $$\left[\lim_{y \to 0} (1+y)^{1/y}\right]^{iz} = e^{iz}.$$ But we don't know that $$e = \lim_{y \to 0} (1+y)^{1/y}.$$ We do know that $$e = \lim_{y \to 0+} (1+y)^{1/y}.$$ For $y \in \mathbb{R}.$ In other words, we know that the right limit of $(1+y)^{1/y}$ is $e$, but I wan't to be able to show that its true for all $y \in \mathbb{C}$ and for all "paths".

$\endgroup$
  • $\begingroup$ Alternatively, you can define power as inverse of logarithm where logarithm is integral of inverse and inverse is $x\times x^{-1}=1$. $\endgroup$ – Arash Dec 2 '17 at 23:48
  • 2
    $\begingroup$ This is meaningless unless you can define what $b^i$ means. $\endgroup$ – fleablood Dec 3 '17 at 0:01
  • $\begingroup$ @fleabood Good point $\endgroup$ – Jbag1212 Dec 3 '17 at 0:41
  • $\begingroup$ My answer is a slight improvement over your attempt above. $\endgroup$ – Wlod AA Dec 19 '17 at 22:30
11
$\begingroup$

We want to define $$f(z)=e^z$$ in such a way that it satisfies the following properties: $$ f'(z) = f(z) , \quad f(x+ 0i) = e^x$$ In other words we want it to be its own derivative and we would like it to reduce to the regular exponential function when the exponent is purely real. Lets explore what properties such a function would have to have.

To make things easier write $f(z) = f(x+iy) = u + iv$ with $ u = u(x,y) , v = v(x,y)$.

By the Cauchy-Riemann equations we know that $$f'(x+iy) = u_x + i v_x = v_y + i (-u_y) = u + iv.$$ where the rightmost equality comes from the fact that$ f' = f$. Equating real/imaginary parts we see that $$u(x,y) = u_x(x,y)$$ $$ v(x,y) = v_x(x,y)$$ for all x,y. The general solutions to these equations are $$u(x,y) = a(y)e^x$$ $$ v(x,y)= b(y)e^x $$ Looking at the other constraint we have $$ e^{x} + 0i = e^ x = f(x+0i) = u(x,0) + i v(x,0) = a(0)e^x + i b(0)e^x$$ This gives our "initial conditions" $$ a(0) = 1 ,\ b(0) = 0.$$

Going back to the C-R equations we have $$ u_x = v_y \implies a(y)e^x = b'(y)e^x$$ $$v_x = - u_y \implies a'(y)e^x = -b(y)e^x$$ Giving the system $$a = b'$$ $$- b = a'$$ Which we can cleverly write as $$ \vec{x}' = \begin{bmatrix} a'(y) \\ b' (y) \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a(y) \\ b(y) \end{bmatrix} = A \vec{x} $$ It turns out the solution to a linear system like this is given by the matrix exponential

$$ \vec{x}(y) = e^{Ay} \vec{x}_0 $$

$ \\ $

where $ \vec{x}_0 = \begin{bmatrix} a(0) \\ b(0) \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $ e^{Ay} = \displaystyle\sum_{k=0}^{\infty} \frac{A^ky^k}{k!} $ and$A^k$ denotes matrix exponentiation.

Note that $ A^2 = -I$ so that $A^3 = - A$, $A^4 = I$, $A^5 = A $ etc.

This gives $$ e^{Ay} = \displaystyle\sum_{k=0}^{\infty} \frac{A^ky^k}{k!} = \displaystyle\sum_{\text{even } k} \frac{A^ky^k}{k!} + \displaystyle\sum_{\text{odd } k} \frac{A^ky^k}{k!}$$

$\\$

$$ = \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k Iy^{(2k)}}{2k!} + \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^kAy^{(2k+1)}}{(2k+1)!}$$ $= \begin{bmatrix} \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!} & 0 \\ 0 & \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!}\end{bmatrix} + \begin{bmatrix} 0 & - \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} \\ \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} & 0 \end{bmatrix} $

$\\$

$= \begin{bmatrix} \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!} & - \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} \\ \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} & \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!} \end{bmatrix}= \begin{bmatrix} \cos(y) & - \sin(y) \\ \sin(y) & \cos(y) \end{bmatrix}$

As mentioned above, multiplying by the initial conditions vector gives us our solution : $$ \begin{bmatrix} a(y) \\ b(y) \end{bmatrix} = \begin{bmatrix} \cos(y) & - \sin(y) \\ \sin(y) & \cos(y) \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} cos(y) \\ sin(y) \end{bmatrix}$$ We finally arrive at $$f(x+iy) = u(x,y) + i v(x,y) = e^x\cos(y) + i e^x\sin(y).$$ In other words, if we want the complex exponential to naturally generalize the real exponential then

we must DEFINE it as $ e^{x+iy} = e^x(\cos(y) + i \sin(y))$.

$\endgroup$
  • $\begingroup$ This is a good answer. Is there any way to make the argument work by "fiddling" with the limits, like I was originally trying to do? $\endgroup$ – Jbag1212 Dec 3 '17 at 0:47
  • 1
    $\begingroup$ @Jbag1212 I think probably not because of lifeblood's comment above. In real analysis, one defines $a^{x}:= e^{x ln(a)}$. I don't remember how it is defined for complex but if its the same way you are using a function thats defined in terms of the complex exponential to derive the complex exponential. Which is a problem. If its not, then perhaps it could, I'm actually quite sick atm, I just basically copy and pasted my answer to another old post like this. Heads not super clear atm. When I have a chance I will let you know with more certainty $\endgroup$ – David Reed Dec 3 '17 at 1:23
  • $\begingroup$ I've made some minor cosmetics changes into your excellent answer. Please let me know if this bothers you. $\endgroup$ – Dmoreno Jan 9 '18 at 8:47
  • $\begingroup$ On the other hand, I think you can make your life easier by differentiating your couple of ODEs for $a$ and $b$, thus producing $a''+a=0$ and $b''+b=0$. Solving these yields the same answer. $\endgroup$ – Dmoreno Jan 9 '18 at 8:52
2
$\begingroup$

Let's simply define $\ \exp(z)\ $ instead of $\ \exp(i\cdot z).\ $ Here:

$$ \exp(z)\,\ :=\,\ \lim_{n\rightarrow\infty}\ \left(1 +\frac zn\right)^n $$

Then, as requested in another Answer, $\ \exp(x+i\cdot 0) = \exp(x)\ $ (of course), and derivative $\ exp\,'\ =\ \exp.$

REMARK. A clean definition (like the one here) should not use irrational exponents nor even any non-integer exponent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.