# $\pi$ in imaginary numbers?

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 $\pi$?

• 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 $0$. – MoebiusCorzer Dec 11 '15 at 13:15
• Do you know De Moivre's formula, $(\cos x+i\sin x)^n=\cos(xn)+i\sin(xn)$? – Akiva Weinberger Dec 11 '15 at 14:09
• 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: $$4 \sum_{k = 1}^\infty \frac{(-1)^{k + 1}}{2k - 1}.$$ – Robert Soupe Dec 11 '15 at 17:46
• In Wolfram Alpha, try 4Sum[((-1)^(k + 1))/(2k - 1), {k, Infinity}] – Robert Soupe Dec 11 '15 at 18:23

## 3 Answers

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 :

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

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.

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