Euler's formula, is this true? [duplicate]

*I've changed this question as below.

Let me have a function such as $f(k) = \exp(j 2 \pi k )$, where $k$ is real value.

Using Euler's formula, we can write $f(k)$ as below,

$$f(k) = \exp(j 2 \pi k ) = \cos(2\pi k)+j \sin(2\pi k).$$

If $k$ is integer, this always goes to 1.

Up to here, nothing is weird and makes sense.

However, if the equation goes to

$$f(k) = \exp(j2\pi k)=e^{j2\pi k}=(e^{j2\pi})^k = (\cos(2\pi)+j\sin(2\pi))^k=1^k=1$$

What I want to know is this above equation makes sense or not.

marked as duplicate by Marc van Leeuwen, dustin, user147263, Claude Leibovici, kingW3Feb 1 '15 at 9:55

• Isn't there a typo ? $f(x)=\exp(j2\pi x)$ makes more sense. – Yves Daoust Jan 31 '15 at 16:45
• Note that it is true that $f(k)=(e^k)^{2\pi\mathbf j}$, for all real $k$. (Also, it would be less confusing to call a real number something like $x$.) – Marc van Leeuwen Feb 1 '15 at 5:59

No, in complex numbers you may not assume that $(e^a)^b=e^{(ab)}$. You just found a counter-example.

• thank you for your reply. But I mentioned that $k$ is real value. By the way, if $k$ is an integer value, above equation makes sense from de Moivre's formula? – Creatlee Jan 31 '15 at 17:02
• with complex numbers, the notation $x^y$ should be avoided, except when $x=e$ – mercio Jan 31 '15 at 17:11
• $k$ integer is a counter-counter-example. – Yves Daoust Jan 31 '15 at 18:14
• You might want to add that the equality in this answer does hold whenever $a\in\Bbb R$. The value of $b$ can be any complex number. – Marc van Leeuwen Feb 1 '15 at 6:06
• @mercio Or when $y$ is an integer! – Bruno Joyal Feb 1 '15 at 6:11

The issue that you are having is that $z^k$ can be a multivalued function, dependent on $k$. As a basic, example, we all know that $(2)^2 = 4 = (-2)^2$. So we can just as well define $4^{1/2}$ as $+2$ or $-2$.

By your argument, we can have (the clearly erroneous result) $$-1 = e^{\pi i} = e^{2 \pi i \cdot {1 \over 2}} = (e^{2 \pi i})^{1/2} = 1^{1/2} = 1.$$ But, as with my example above, but using $(1)^2 = 1 = (-1)^2$, we see that we can define $1^{1/2}$ as either $+1$ or $-1$. The technical complex analysis term is a 'branch cut' - using one of these allows you to define $z^k$ uniquely.

I've answered a question before on a similar topic - Why do I get two different results for the reciprocal of i?.

Hopefully this answer has been helpful to you! If it has, then please remember to upvote and/or accept! Hope you understand now! :)

• Multivalued functions is a different topic. In the question, the base is a positive real number, and exponentiation is well defined. $4^{1/2}=\sqrt4=2$, period. – Marc van Leeuwen Feb 1 '15 at 6:03
• I think the point is that the original is, in general, a complex number, and exponent certainly is a complex number. As you've mentioned in your answer, you can "take the power out" when you leave the real power inside. Otherwise you have to consider branch cuts. My linked example was supposed to be a similar example, not a possible duplicate. :) – Sam T Feb 1 '15 at 9:11
• In the context of complex numbers, however, you can certainly define $4^{1/2} = -2$. I could define the square root so that, writing $\sqrt{\cdot}$ for the positive square root of a real number, $$1^{1/2} = -1, \ \ 3^{1/2} = -\sqrt{3}, \ \ 5^{1/2} = \sqrt{5}$$ and so on, should I so desire. Just take my branch cut to be $(t,\sin({1 \over 2} \pi t)) \in \Bbb C$. Would be rather obtuse to do so! =P – Sam T Feb 1 '15 at 9:19
• You can define whatever pleases you most, but this (kind of) question is about the value to attribute (or not) to certain arithmetic expressions, not some advanced study of complex functions. Expressions serve to denote (single) values, when you take that away from them, you are robbing them of their principal utility. – Marc van Leeuwen Feb 1 '15 at 10:00
• But isn't that actually the whole point of this question? The mistake comes from the fact that you can't just say $\exp( 2 \pi ik ) = (e^{ 2 \pi i} )^{k}$? I thought that this was the reason; am I mistaken? (You've got 54k rep and I've got about 700!) – Sam T Feb 1 '15 at 10:30

Following on from Yves Daoust:

$$(z^x)^y$$ $$=(e^{log(z^x)})^y=(e^{xlog(z)})^y$$ $$=e^{log((e^{xlog(z)})^y)}=e^{ylog(e^{xlog(z)})}$$ $$= e^{y(xlog(z)+i(2n\pi))}$$ $$=z^{xy} \cdot e^{i(2ny\pi)}$$

• Thank you for your reply. By the way, what I want to know is simple. In your equation also, there is $exp \{i (2ny \pi) \}$. I think this equals 1 without considering $ny$. Because $exp (i2 \pi)$ is always 1. $1 ^{ny}$ is always 1. – Creatlee Jan 31 '15 at 17:33
• @AlbertRee You have to consider the $y$ in the exponent. While $n$ is an integer, $y$ is not necessarily an integer or even real. Try computing $e^{i2y\pi}$ for $y=\frac12 , i , \frac{-i}{3}$. You won't get the same value each time. The laws of indices do not all work in the same way for complex numbers as they do for reals. – Gridley Quayle Jan 31 '15 at 17:48

For non-integer exponents, exponentiation $x^y$ is only well defined (denotes a single value) when $x$ is a positive real number; on the other hand $y$ can be any complex number. (And for most purposes one can do with just using $x=e=\exp(1)$, but that is a separate matter). Assuming that, the rule $x^{y+z}=x^yx^z$ is always valid. However the rule for multiplication in the exponent has a restriction that is rarely stated explicitly: $$x^{yz}=(x^y)^z \qquad \text{provided that x>0 is real} \textbf{ and that } \text{y is real.}$$ Note that it does not suffice that the expressions on both sides are defined, in other words that $x$ and $x^y$ are positive real numbers: both sides can be well defined value, yet different. The example in the question illustrates this (for any non-integer value of$~k$).

See this answer to a similar question for more details, and a proof of the rule stated above.