Does Euler's formula remain valid if we use any real number other than the constant $e$? For example replacing $e$ with 5 would make the formula look like this: $5^{it}$.

I tried this idea in Matlab and replaced $e$ with few other real numbers (eg 1.5, 10, 2.1) and each time the plot was still showing what seemed like cosine and sine waves. The frequency of the cos and sin was changing depending on the base.

Here's roughly my approach:

w = freq * 2 * pi;
t = 0:0.001:1000 ;

a = real( number ^ (i*wt) ) ; % cos in Euler's formula
b = imag( number ^ (i*wt) ) ; % sin in Euler's formula

 Example plot of real and imaginary components of: 1.5^(i*2*pi*100*t)


3 Answers 3


Say you're interested in $$M^{j2\pi f_0 t}. \tag{1}$$ Note that $$M = e^{\log M},$$ so $(1)$ can be written as

\begin{align} M^{j2\pi f_0 t} &= \left( e^{\log M} \right) ^ {j2\pi f_0 t} \\ &= e^{j2\pi (f_0\log M) t} \\ &= \cos(2\pi (f_0\log M) t) + j \sin(2\pi (f_0\log M) t), \end{align} which is a complex sinusoid with frequency $f_0 \log M$. That is why using $M$ instead of $e$ results in a change in frequency.


It's an interesting question. Let's see what nonzero complex numbers $w$ have the property that they "act like $e$" in the classical formula, i.e., that $$e^z = w^z$$ for all complex $z=x+iy$. For convenience, suppose we can write $$w=re^{it}$$

The symbol $w^z$ takes the possible multiple values $$w^z =e^{z\log w} = e^{(x+iy)(\overbrace{\ln r + it + 2k\pi i}^{\textrm{possible values of $\log w$}})} =e^{(x\ln r - yt-2k\pi y) +i(y\ln r + xt +2k\pi x)}$$

This means we will have $e^z=w^z$ when $$(x+yi)-[(x\ln r - yt-2k\pi y) +i(y\ln r + xt +2k\pi x)] = 2\pi n i$$ for some $n$. But this means (by equating real and imaginary parts on both sides) $$\begin{cases}x=x\ln r - yt-2k\pi y\\y=y\ln r + xt +2k\pi x + 2\pi n \end{cases}$$ This can happen for all $z$ (i.e., all $x,y$) only if $r=e$ and $t=k=n=0$.

But that means $w =e\cdot e^{0i}=e$, so there is no other complex number $w$ that will do the trick.


For any a, $a= e^{ln(a)}$ because "$e^x$" and "ln(x)" are "inverse functions. So $$a^{it}= e^{\ln(a^{it})}= e^{it \ln(a)}$$. Then $$a^{it}= e^{i (t \ln(a))}= \cos(t \ln(a))+ i \sin(t \ln(a)).$$

  • $\begingroup$ For positive $a$ $\endgroup$ Commented Oct 5, 2017 at 21:22
  • $\begingroup$ @HallsofIvy : This is not entirely correct. Even assuming $a>0$, $a^{it}$ takes multiple values: $$a^{it}=e^{it(\ln a + 2\pi ki)} = e^{-2\pi kt + it\ln a} = e^{-2\pi kt}(\cos(t\ln a) + i\sin(t\ln a))$$ (taking $k=0$ recovers your specific value). If $a$ is negative, or not real, it's even more complicated. $\endgroup$
    – MPW
    Commented Oct 18, 2017 at 12:45

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