If e^(2ipi) = 1, and e^0 = 1, can I say that 2ipi = 0?

Question

I know I can't, but I really don't understand why. Also, please don't give me the background of Euler's formula.

Answer 1

In a weird and strange way, yes you can say that just like how you can say that 0=2pi=4pi=6pi=...because sin(0)=sin(2pi)=(sin4pi)=... How can we say that 0=2pi=4pi=6pi=..., we can say that because when we go around a circle centered at zero, zero radians is that same as 2pi radians. If I start at zero and I move 2pi radians, I end up at the same point so I can refer to it as either zero radians or 2pi radians.

Stricly spekaing, 2ipi isnot zero because the complex numbers are in integral domain. There are no zero divisors meaning if I have a bunch of complex numbers multiplying out to zero, at least one of them has to be zero. 2 is nonzero, i is non zero, pi is nonzero, their product cannot be zero.

Contrast this example with the set of 2x2 matrices. You can easily have two nonzero matrices multiplying out to the zero matrix.

[{3,5},{3,5}]*[{5,-10},{-3,6}]
=[{0,0},{0,0}]

The reason you argument fails, is that you are taking the natural log of both sides, right? You cannot do that because for precisely this reason, the exponential is not a one-to-one function in the complex plane. It is one-to-one for the real numbers but in the complex plane, it is not. It is periodic. Therefore, no such inverse exists. If you want to define the inverse, then you have to restrict the domain and range to make the exponential one-to-one and onto.

The same thing happens in the real numbers with the sine function for example. y=sin(x) is a function but x=sin(y) is not a funtion in f. if we want the arcsin, then we have to restrict the domain to -pi to pi for example which fixes the range from 1 to -1.

In other words, when using don't complex numbers, don't take their log.

BTW, zero is a complex number also, written as 0=0+0i.

Answer 2

The reason you can't do this is perhaps best explained by considering a simpler version of the same phenomenon:

(-1)^2=1^2, -1≠1.

The reason this equation does not imply that -1 = 1 is that squaring two different numbers can indeed give you the same answer. In general, simply because f(x) = f(y), does not imply that x=y. There are some functions for which f(x)=f(y) does imply that x=y, and these functions are called one-to-one. This includes some of the more common functions, such as addition or subtraction of any constant, and multiplication or division by any nonzero constant (a fact which makes algebra possible). However, as you have discovered, it does not include e^x on the complex plane. For this reason, the complex logarithm is considered to be multivalued, since there are many numbers which, when e is raised to their power, yield any particular number.

Note that e^x = e^y does imply a weaker equality, namely that x = y ± 2kπi, for some integer k.

Answer 3

No. The only way that the product of numbers in the complex plane can be zero is when one of them is zero. Now, 2, i, pi, are all non zero.

To solve e^(2ipi)=e^0 when you take log of both sides you need to take in account the argument of the angle since you are in the complex numbers, so you have something extra on the left hand side.

Answer 4

No, you can't, and you have some long-winded explanations already.

Here's a simpler one: if a^x = a^y, then all you can say is that a^(y-x) = 1.

This equation may look trivial, but it isn't. When a, x and y are integers and the exponentiation is in MODULAR arithmetic, it is this very identity which underpins all encryption on the Internet.

Answer 5

hmmm...probable...however the single that suits terrific for the wizard or the single he's terrific suited for could be the single chosen. it is not likely although that the point of compatibility between the two wands would be realtively extreme! One is for specific to be extra well suited. additionally, each wand has a center, and a definite lentgh. The length of a wand relies upon on the scale of the wizards arm, and the middle finally relies upon on what form of magic the wizard prefers. If a wizard is extra apt for Transfiuration , as an occasion, then that individual would be extra constructive to a definite middle. and additionally of direction if the wand is gained, then that wand provides you with its allegiance...making you an proprietor of two wands...very corresponding to Harry and Dumbledore! So i think of its obtainable, yet fantastically no longer likely! And it very hardly occurs (i think of)

Answer 6

no,because 2ipi is imaginary and cannot be equated to 0 which is real