I am confused about something: \begin{eqnarray} (e^{2 i \pi})^{0.5} = (e^{2 i \pi \cdot 0.5})= e^{i \pi}=-1 \end{eqnarray} but \begin{eqnarray} e^{2 i \pi}=1~ and~ 1^{0.5}=1 \end{eqnarray} where is my mistake??
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4The short version: if $z$ is complex, then $z^{0.5}$ is meaningless. – Jack M Jan 15 '15 at 00:35
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4Note that just like every positive real number, 1 has two square roots: 1 and -1. – BHT Jan 15 '15 at 00:37
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4You have discovered multi-valued roots; but you knew about them already, right? Both +1 and -1 are square roots of +1. – Joffan Jan 15 '15 at 00:37
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1^0.5=-1? I dont understand this – A_l Jan 15 '15 at 00:38
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1$\sqrt{1} = \pm 1$. (Well, the real roots, anyway!) Also, $(-1)^2 = 1^2 = 1$. – John Jan 15 '15 at 00:44
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Slightly longer version: If $z$ is complex, then $z^{0.5}$ is not a function. – Matthew Leingang Jan 15 '15 at 00:45
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@MatthewLeingang, wouldn't that be true even over the reals? I mean $1^{.5} = \pm 1$, right? Or do we just define exponents as the positive root over the reals? – k_g Jan 15 '15 at 05:42
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1@k_g: Yes, it's true over the reals, too. The way out is that we restrict $x$ to nonnegative real numbers in order to make $\sqrt{x}$ a function. But with complex numbers it's customary to consider $\sqrt{z}$ a multi-valued function. – Matthew Leingang Jan 15 '15 at 09:41
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@JackM: Complaining that $z^{1/2}$ is meaningless for (non-real) complex $z$ is not to the point, because $\exp(2\pi\mathbf i)$ is actually real. – Marc van Leeuwen Jan 15 '15 at 10:30
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@MatthewLeingang I've had profs who'd insist that $ e^{2 \pi i} $ is complex, and happens to be a complex number with a zero imaginary part. This context here is one where the distinction matters. – greggo Jan 15 '15 at 14:31
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@JackM may have been pointing out, tersely, that while $z^{\frac{1}{2}}$ has two values, $z^{0.5}$ is meaningless. – greggo Jan 15 '15 at 14:34
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Another related post: What is wrong with this fake proof $e^i = 1$? – Martin Sleziak Aug 30 '19 at 12:20
4 Answers
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It's exactly as if you would write:
I am confused about something: $$((-1)^2)^{0.5} = (-1)^{2\cdot 0.5} = (-1)^1 = -1$$ but $$(-1)^2=1 \text{ and } 1^{0.5} = 1$$ where is my mistake??
Ystar
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This is not the same situation, because in the question everything that appears as base of an exponentiation is actually a positive real number, unlike $-1$ in this answer. – Marc van Leeuwen Jan 15 '15 at 10:34
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When dealing with complex numbers, you no longer interchange radicals with fractional exponents, in the same way you did with positive real numbers.
If $x$ is real, $\sqrt[3]x$ denotes the principal cube root of $x$, and has only one value. In this way $8^{1/3}$ and $\sqrt[3]{8}$ mean the same thing.
But if we are doing algebra with complex numbers, and we wrote $5^{1/3}$, we would mean any solution to $z^3 - 5 = 0$.
So $1^{1/2}$ is defined as any solution to $z^2 - 1 = 0$. $z = 1$ and $z = -1$ are both solutions to this.
GFauxPas
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Would you also say that $e^{0.5}$ could stand for the negative number $-\sqrt e,$? – Marc van Leeuwen Jan 15 '15 at 11:07
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Exponentation $x^y$ can be unambiguously defined whenever $x$ is a positive real number, and $y$ is any complex number. The value is (per definition if you like) equal to $\exp(y\ln(x))$ where $\exp:\Bbb C\to\Bbb C$ is the usual exponential function, which is well defined and well behaved. [Writing it using $e^z$ would give a circular definition here.] But the rule $x^{yz}=(x^y)^z$ is only valid when $y$ is real; this is more strict than requiring $x^y$ to be real (as the question illustrates), which is the condition necessary for $(x^y)^z$ to be defined in the first place.
The proof that $x^{yz}=(x^y)^z$ when $x>0$ and $y$ are real is simple: $$ x^{yz}\stackrel{\rm def}=\exp(yz\ln(x)) = \exp(z\ln(x^y))\stackrel{\rm fed}=(x^y)^z \qquad(x,y\in\Bbb R, x>0, z\in\Bbb C). $$ This also shows that that a situation with $x^{yz}\neq(x^y)^z$ where both sides are defined but $y\notin\Bbb R$ must boil down to a similar but more basic situation involving $\ln$. Indeed the rule $y\ln(x)=\ln(x^y)$ is not valid for $y\notin\Bbb R$ even if the arguments to $\ln$ are both real: $$\def\ii{\mathbf i} 2\pi\ii=2\pi\ii\ln(e)\neq\ln(e^{2\pi\ii})=\ln(1)=0. $$
Curiously, the failure of $x^{yz}=(x^y)^z$ when $y\notin\Bbb R$ does not dissuade people to employ this very rule when trying to "define" such exressions as $\ii\,^\ii$. I find such attempts truly pathetic. There is no point in trying to define exponentiation beyond the case of positive real base, or integer exponent; one gains nothing but confusion.
Marc van Leeuwen
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+1 I was going to add that complex to the rational power $z^{n/d}$ is not meaningless, albeit having $d$ values, but on further thought about what you wrote, I see that such cases are much better stated as e.g. $\sqrt[d]{z^n}$, avoiding confusion. – greggo Jan 15 '15 at 15:00
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@GFauxPas: Because $\pi$ is a positive real number, but $\mathbf i$ is not. There is not natural sense in which raising $\mathbf i$ to a non-integral power means anything. – Marc van Leeuwen Jan 15 '15 at 20:07
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2"Multiplying $\pi$ by itself $\pi$ times" doesn't have an intuitive meaning either, though. But this is a derailment. – GFauxPas Jan 15 '15 at 21:10
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4$\pi^\pi$ may be a strange number that you would never need, but it has a clear meaning. Take any exponential growth process; mark the time it takes to grow by a factor $\pi$ and then let it evolve $\pi$ times as long. – Marc van Leeuwen Jan 16 '15 at 05:38
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Here, let me try and make an analogy that you might be able to understand simply. As stated in the comments, 1 has two square roots: 1 and -1. See why this is important below.
$$((-1)^{2})^{0.5} = (-1)^{(2 \times 0.5)} = (-1)^1 = -1$$
OR
$$((-1)^{2})^{0.5} = 1^{0.5} = \sqrt{1} = 1$$
Do you get it?
louie mcconnell
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