7
$\begingroup$

Look at the result of $(-1)^{1/10000000}$ on the google calculator. You should get $$1 + 3.14159265 \times 10^{-7} i$$ Why does $\pi$ occur in imaginary number operations that don't include $\pi$?

$\endgroup$
  • 1
    $\begingroup$ Firstly, your calculation is an approximation of $\pi$. Secondly, you may want to look at this page explaining $e^{2i\pi}+1=0$ Also, look at this and be aware of the fact that $\sin(x)\approx x$ when $x$ is close to $0$. $\endgroup$ – MoebiusCorzer Dec 11 '15 at 13:15
  • $\begingroup$ Do you know De Moivre's formula, $(\cos x+i\sin x)^n=\cos(xn)+i\sin(xn)$? $\endgroup$ – Akiva Weinberger Dec 11 '15 at 14:09
  • 1
    $\begingroup$ Most of us know about $\pi$ from geometry, but it has many manifestations in number theory. Peruse mathworld.wolfram.com/PiFormulas.html Of particular interest to you: $$4 \sum_{k = 1}^\infty \frac{(-1)^{k + 1}}{2k - 1}.$$ $\endgroup$ – Robert Soupe Dec 11 '15 at 17:46
  • $\begingroup$ In Wolfram Alpha, try 4Sum[((-1)^(k + 1))/(2k - 1), {k, Infinity}] $\endgroup$ – Robert Soupe Dec 11 '15 at 18:23
14
$\begingroup$

You know that $e^{i\pi}=-1$. Then :

$$(-1)^{\frac{1}{10^7}}=e^{\frac{i\pi}{10^7}} $$

since $x:=\frac{i\pi}{10^7}$ is very small you can approximate $e^x$ by $1+x$ hence :

$$(-1)^{\frac{1}{10^7}}=e^{x} \approx 1+x = 1+\pi.10^{-7}i $$

$\endgroup$
2
$\begingroup$

Let me refer to the Euler's formula $$e^{ix} = \cos x + i \sin x.$$ There are many ways to 'derive' the formula but I am not going to make the derivation. You may find its nice derivations from Wikipedia.

In some textbook they define exponential function $e^x$ as other ways (for example, infinite series) and provide its inverse $\ln x$. If we have the exponential function and logarithmic function, we can define arbitrary exponent as $$a^x = e^{x \ln a}.$$ We hope that such definition works for complex numbers. Unfortunately, a technical problem arises: complex logarithm is not well-defined.

You can check that $e^{i\pi} = e^{3i\pi} = -1$. It says that the possible values of $\ln(-1)$ are $\pi i$ and $3\pi i$. In fact, $e^z = e^{z+2n\pi i}$ for all integer $n$ and the complex exponential has a period $2\pi i$. That problem is easily resolved: although the exponential function $\exp : \Bbb{C}\to\Bbb{C}-\{0\}$ is not one-to-one (so we can not find its inverse), its restriction on the set of complex $z$ with $-\pi < \operatorname{Im}z \le \pi$ is bijective. Let us call the inverse of it $\operatorname{Ln} z$, and many calculators adopt such definition of complex logarithm.

Now we can define the complex exponential as $$a^z := e^{x\operatorname{Ln} a}.$$ Especially we get $(-1)^z = e^{\pi i z}$. For small real $t$, $$(-1)^t = \cos (\pi t) + i \sin (\pi t) \approx 1 + i\pi t$$ (To derive last approximation, consider the Taylor expansion of $\cos$ and $\sin$.) This is the reason you got such computation result.

$\endgroup$
1
$\begingroup$

$$\left(-1\right)^{\frac{1}{10^7}}=\left(|-1|e^{\arg(-1)i}\right)^{\frac{1}{10^7}}=\left(e^{\pi i}\right)^{\frac{1}{10^7}}=e^{\frac{\pi i}{10^7}}=$$ $$\cos\left(\frac{\pi}{10^7}\right)+\sin\left(\frac{\pi}{10^7}\right)i\approx 0.999+3.1415\cdot10^{-7}i$$

$\endgroup$

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.