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Some of you have probably heard of the famous Euler's Identity, \(e^{iπ}+1=0\). But for those of you who don't know where this came from, have you ever wondered? Well, as seen in this question's name, it comes from **Euler's Formula.**

Euler's formula, named after Leonhard Euler, is pretty simple: \(e^{ix}=\cos x+i\sin x\). Now, don't worry, even if there are trigonometric functions in this equation doesn't mean it's complicated. This formula has applications in complex number theory, trigonometry, algebraic topography, etc. But I'll mostly be focusing in complex analysis for this question. Also, we won't be going into its history; that's what Wikipedia's for. Anyway, back to real math.

Euler's formula in complex analysis is usually a way of converting cartesian coordinates to polar, and vice versa.

**QUICK CRASH COURSE ON CARTESIAN AND POLAR COORDINATE SYSTEMS!**

Cartesian coordinates are coordinates, well, on the cartesian plane. Your garden variety (x, y) ordered pairs and whatnot. Polar coordinates are a type of coordinates where the angle in radians and the distance from a reference point is measured in the form (r, \(\phi \)), where r is the distance and \(\phi\) is the angle. Got that all in your head? Good, we can proceed. |

But before we tackle Euler's formula, we must take a look at polar coordinate conversion. Polar coordinates r and \(\phi\) can be converted to Cartesian coordinates as follows:

\(x=r\cos \phi\), \(y=r\sin \phi\), \(r=\sqrt{x^2+y^2}\) using Pythagoras' Theorem, and \(\phi=\arctan2(y,x)\) wherein arctan2 is a variation of the arctan function. I won't go into this. |

To convert complex numbers (\(a+bi\)) into polar coordinates, we have to turn to Euler's formula. The formula for converting any complex number or its conjugate (in a nutshell, what to multiply to it to get rid of that pesky \(i\)) is

\(z=a+bi=r*\cos \phi+r*i\sin \phi=r(\cos \phi+i\sin \phi)\), and \(\overline z=a-bi=r*\cos \phi-r*i\sin \phi=r(\cos \phi +i\sin\phi)\) wherein \(z\) is our point of interest, and \(\overline z\) is z's complex conjugate |

As we can see in the box above, we have – lo and behold – Euler's formula! Therefore we can simplify this further, and when we do this, we get \(z=re^{i\phi}\)! Neat! Now we have a good-looking formula for converting polar form coordinates into Cartesian/rectangular form.

This is great and all, but what else? Where are the other more fun uses to this formula besides circular coordinate systems? Well, for example, it can be used to find the natural logarithm of \(i\).

Using Euler's Formula, \(e^{i{\pi \over 2}}=\cos {\pi \over 2}+i\sin {\pi\over 2}=0+i(1)=i\). Getting the \(\ln\) of both sides of the equation, we get the following: \(\ln(e^{i{\pi\over 2}})=\ln i\), which, simplified, becomes \(i{\pi\over 2}=\ln i\)! This is great! |

You can redefine the \(\sin\) and \(\cos\) functions in a few easy steps:

\(e^{ix}=\cos x+i\sin x\) \(e^{-ix}=\cos(-x)+i\sin(-x)=\cos x-i\sin x\) \(e^{ix}+e^{-ix}=2\cos x\) \({e^{ix}+e^{-ix}\over 2}=\cos x\) \(e^{ix}-e^{-ix}=2i\sin x\) \({e^{ix}-e^{-ix}\over 2i}=\sin x\) |

In fact, it can even lead to some evil math.

\(e^{i(2\pi)}=\cos 2π+i\sin 2π=1\). But we also know that \(e^0=1\). Using the transitive property of equality, we get the following equation: \(e^{i2π}=e^0\). Getting the \(\ln\) of both sides, we get \(2iπ=0\). Wait. this isn't right... This means \(2=0\)? And \(i=0\)? And... \(\pi=0\)?! Okay. Something's definitely wrong here. Try to find the fallacy! |

That one above, I found it by playing around . Anyway, as you can see here, Euler's formula can get pretty... entertaining, to say the least. You can do much more with this formula if you play around with it. Now, for what some of us came for; how do we get \(e^{iπ}+1=0\)? Well, now that we know what Euler's formula is, getting this should be extremely easy!

If we subtitute \(π\) into Euler's formula, we get \(e^{iπ}=\cos π+i\sin π=-1+i(0)=-1\). Adding 1 to both sides, we get \(e^{iπ}+1=0\)! |

You see, simple. Credits to Leonhard Euler for coming up with the formula, for without it, this question would not exist. Anyway, as always, I'll leave you guys with a (pretty simple) challenge:

Find \(x\) such that \(e^{ix}=e^{i2x}\) EDIT: and \(x≠0\) (sorry hectictar)

Good luck, and thanks for reading.

Mathhemathh Oct 3, 2017

#2**0 **

Solve for x:

e^(2 i x) = e^(i x)

e^(2 i x) = e^(2 i x) and e^(i x) = e^(i x):

e^((2 i) x) = e^(i x)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a). Assume the original equation as extraneous roots may be introduced:

(2 i) x = i x + (2 i) π n for n element Z and e^(2 i x) = e^(i x)

Subtract i x from both sides:

i x = (2 i) π n for n element Z and e^(2 i x) = e^(i x)

Divide both sides by i:

x = 2 π n for n element Z and e^(2 i x) = e^(i x)

The roots x = 2 π n never violate e^(2 i x) = e^(i x), which means this assumption can be omitted:

**x = 2 π n for n element Z**

Guest Oct 4, 2017