注：本文为 “虚数
i
i
i ” 相关文章合辑。
机翻，未校。
i to the power i
2023-08-08 00:00:00
The imaginary unit i i i is defined to be the positive square root of -1. But what is i i i to the power i i i? Is it even possible to calculate, and what does it mean?
虚数单位 i i i 被定义为 − 1 -1 −1 的正平方根。但是 i i i 的 i i i 次幂是多少呢？它是否可以计算，又意味着什么呢？
As we will see, it is possible to calculate i to the power i, and the result is quite surprising in a couple of different ways. But we will start with a quick recap on the real powers of complex numbers, in particular the real powers of i.
正如我们将看到的，计算 i i i 的 i i i 次幂是可能的，并且结果在几个不同方面相当令人惊讶。但我们将从快速回顾复数的实幂开始，特别是 i i i 的实幂。
Modulus-argument form for multiplication
We will be using the modulus-argument form for complex numbers, where a complex number z is represented as a radius r (called the modulus) and an angle Θ (called the argument):
我们将使用复数的模 - 辐角形式，其中复数
z
z
z 表示为半径
r
r
r （称为模）和角度
θ
\theta
θ （称为辐角）：
z = r e i θ \Large z = r e^{i\theta} z=reiθ
The modulus of z is the distance from the origin to the point z on an Argand diagram. The argument of z is the angle z makes with the x-axis:
z
z
z 的模是复平面上从原点到点
z
z
z 的距离。
z
z
z 的辐角是
z
z
z 与
x
x
x 轴所成的角度：
When we multiply two complex numbers z1 and z2 that are expressed in this form, the normal rules of the exponential function apply:
当我们将两个以这种形式表示的复数
z
1
z_1
z1 和
z
2
z_2
z2 相乘时，指数函数的常规规则适用：
z 1 z 2 = r 1 e i θ 1 ⋅ r 2 e i θ 2 = r 1 r 2 e i ( θ 1 + θ 2 ) \Large z_1 z_2 = r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)} z1z2=r1eiθ1⋅r2eiθ2=r1r2ei(θ1+θ2)
We multiply the moduli r1 and r2. We add the arguments Θ1 and Θ2. That is exactly the same as we would do if the exponents were real numbers.
我们将模
r
1
r_1
r1 和
r
2
r_2
r2 相乘，将辐角
θ
1
\theta_1
θ1 和
θ
2
\theta_2
θ2 相加。这与指数为实数时的做法完全相同。
The value i in modulus-argument form
We will be using i quite a lot, so it is useful to know its modulus-argument form. Here is i on an Argand diagram:
我们会经常使用
i
i
i ，所以知道它的模 - 辐角形式很有用。这是复平面上的
i
i
i ：
i is 1 unit vertically above the origin. So the length r is 1, and the angle is
π
2
\frac{\pi}{2}
2π radians (which is 90 degrees of course). Here is the exponential form of i:
i
i
i 在原点垂直上方
1
1
1 个单位处。所以长度
r
r
r 为
1
1
1 ，角度为
π
2
\frac{\pi}{2}
2π 弧度（当然也就是
90
90
90 度）。这是
i
i
i 的指数形式：
i = e i π 2 \Large i = e^{i\frac{\pi}{2}} i=ei2π
If we multiply any number z by i, then in modulus-argument form this is:
如果我们将任何数
z
z
z 乘以
i
i
i ，那么在模 - 辐角形式下这是：
z ⋅ i = r e i θ ⋅ e i π 2 = r e i ( θ + π 2 ) \Large z \cdot i = r e^{i\theta} \cdot e^{i\frac{\pi}{2}} = r e^{i(\theta + \frac{\pi}{2})} z⋅i=reiθ⋅ei2π=rei(θ+2π)
In other words, multiplying by z by i simply rotates z by
π
2
\frac{\pi}{2}
2π radians about the origin.
换句话说，将
z
z
z 乘以
i
i
i 只是将
z
z
z 绕原点旋转
π
2
\frac{\pi}{2}
2π 弧度。
Integer powers of i
Before calculating
i
i
i to the power
i
i
i, it is worth looking at
i
i
i raised to a real power, as this will give us a couple of insights into the problem. We can calculate i squared like this:
在计算
i
i
i 的
i
i
i 次幂之前，值得看看
i
i
i 的实次幂，因为这会给我们一些关于这个问题的见解。我们可以这样计算
i
i
i 的平方：
i 2 = e i π 2 ⋅ e i π 2 = e i π \Large i^2 = e^{i\frac{\pi}{2}} \cdot e^{i\frac{\pi}{2}} = e^{i\pi} i2=ei2π⋅ei2π=eiπ
This value has a unit length and an angle of
π
\pi
π radians (half a full turn). This makes it equal to -1. But we already know that i squared is -1, by definition. So (as expected) the modulus-argument form of i squared gives the same result as simple complex number multiplication.
这个值的长度为
1
1
1 ，角度为
π
\pi
π 弧度（半圈）。这使得它等于
−
1
-1
−1 。但根据定义，我们已经知道
i
i
i 的平方是
−
1
-1
−1 。所以（如预期的那样）
i
i
i 平方的模 - 辐角形式给出的结果与简单的复数乘法相同。
We can find i cubed in the same way. This time the angle is
3
π
2
\frac{3\pi}{2}
23π radians (three-quarters of a full turn), so the result is -i:
我们可以用同样的方法找到
i
i
i 的立方。这次角度是
3
π
2
\frac{3\pi}{2}
23π 弧度（四分之三圈），所以结果是
−
i
-i
−i ：
i 3 = e i π 2 ⋅ e i π 2 ⋅ e i π 2 = e i 3 π 2 \Large i^3 = e^{i\frac{\pi}{2}} \cdot e^{i\frac{\pi}{2}} \cdot e^{i\frac{\pi}{2}} = e^{i\frac{3\pi}{2}} i3=ei2π⋅ei2π⋅ei2π=ei23π
i to the fourth has an angle of
2
π
2\pi
2π radians a full turn), so the result is 1:
i
i
i 的四次方的角度是
2
π
2\pi
2π 弧度（一圈），所以结果是
1
1
1 ：
i 4 = e i π 2 ⋅ e i π 2 ⋅ e i π 2 ⋅ e i π 2 = e i 2 π \Large i^4 = e^{i\frac{\pi}{2}} \cdot e^{i\frac{\pi}{2}} \cdot e^{i\frac{\pi}{2}} \cdot e^{i\frac{\pi}{2}} = e^{i2\pi} i4=ei2π⋅ei2π⋅ei2π⋅ei2π=ei2π
Here are i and its second, third and fourth powers plotted on an Argand diagram:
这是
i
i
i 及其二次、三次和四次幂在复平面上的绘制：
It is no great surprise that i to the fourth power is 1. i to the fourth is just i squared then squared again, and since i squared is -1 then we would expect i to the fourth to be 1.
i
i
i 的四次方是
1
1
1 并不奇怪。
i
i
i 的四次方就是
i
i
i 的平方再平方，因为
i
i
i 的平方是
−
1
-1
−1 ，所以我们预期
i
i
i 的四次方是
1
1
1 。
i 4 = e i π 2 ⋅ e i 3 π 2 ⋅ e i π ⋅ e i π 2 = e i 2 π \Large i^4 = e^{i\frac{\pi}{2}} \cdot e^{i\frac{3\pi}{2}} \cdot e^{i\pi} \cdot e^{i\frac{\pi}{2}} = e^{i2\pi} i4=ei2π⋅ei23π⋅eiπ⋅ei2π=ei2π 。
We can generalise this and say that i to any integer power is equal to:
我们可以推广这个结论，说
i
i
i 的任何整数次幂等于：
i n = e i n π 2 \Large i^n = e^{in\frac{\pi}{2}} in=ein2π
Using this we can find the fifth, sixth and seventh powers on the Argand diagram:
使用这个公式，我们可以在复平面上找到
i
i
i 的第五、第六和第七次幂：
Higher integer powers of i continue rotating round and round the unit circle.
i
i
i 的更高整数次幂继续绕单位圆旋转。
Takeaways 要点
There are two important takeaways from this. The first is that raising i to the power n, in modulus-argument form, works in the same way as raising any other exponential to a power n. We just multiply the exponent by n:
从这里有两个重要的要点。第一个是，将
i
i
i 的
n
n
n 次幂，在模 - 辐角形式下，与将任何其他指数的
n
n
n 次幂的计算方式相同。我们只需将指数乘以
n
n
n ：
( e i θ ) n = e i n θ \Large (e^{i\theta})^n = e^{in\theta} (eiθ)n=einθ
The second is that there are infinitely many ways to express i in modulus argument form. Since i to the fourth is equal to 1, it follows that:
第二个是，有无限多种方式来表示
i
i
i 的模 - 辐角形式。因为
i
i
i 的四次方等于
1
1
1 ，所以：
i = i 5 = i 9 = i 13 ⋯ \Large i = i^5 = i^9 = i^{13} \cdots i=i5=i9=i13⋯
In modulus-argument form:
在模 - 辐角形式下：
e i π 2 = e i 5 π 2 = e i 9 π 2 = e i 13 π 2 ⋯ \Large e^{i\frac{\pi}{2}} = e^{i\frac{5\pi}{2}} = e^{i\frac{9\pi}{2}} = e^{i\frac{13\pi}{2}} \cdots ei2π=ei25π=ei29π=ei213π⋯
In fact, for any complex number z with argument Θ, if we add an integer multiple of
2
π
2\pi
2π to Θ, we will get the same number. This follows from Euler’s formula:
事实上，对于任何辐角为
θ
\theta
θ 的复数
z
z
z ，如果我们给
θ
\theta
θ 加上
2
π
2\pi
2π 的整数倍，我们将得到相同的数。这是由欧拉公式得出的：
e i θ = cos θ + i sin θ \Large e^{i\theta} = \cos\theta + i\sin\theta eiθ=cosθ+isinθ
Adding a multiple of
2
π
2\pi
2π to the angle does not change the value of the sine or cosine functions, because those functions are periodic with period
2
π
2\pi
2π , so:
给角度加上
2
π
2\pi
2π 的倍数不会改变正弦或余弦函数的值，因为这些函数的周期是
2
π
2\pi
2π ，所以：
cos ( θ + 2 n π ) + i sin ( θ + 2 n π ) = cos θ + i sin θ \Large \cos(\theta + 2n\pi) + i\sin(\theta + 2n\pi) = \cos\theta + i\sin\theta cos(θ+2nπ)+isin(θ+2nπ)=cosθ+isinθ
Integer roots of i
So what is the square root of i? Well, the square root of a real number x is given by raising x to the power one-half. What happens if we try the same thing with i?
那么
i
i
i 的平方根是多少呢？嗯，实数
x
x
x 的平方根是通过将
x
x
x 提升到二分之一次幂得到的。如果我们对
i
i
i 尝试同样的事情会发生什么呢？
i = ( e i π 2 ) 1 2 = e i π 4 \Large \sqrt{i} = (e^{i\frac{\pi}{2}})^{\frac{1}{2}} = e^{i\frac{\pi}{4}} i =(ei2π)21=ei4π
But remember that i can also be written as i to the power 5. If we take the square root of this alternate form we get a second square root:
但是记住
i
i
i 也可以写成
i
i
i 的
5
5
5 次幂。如果我们对这个替代形式取平方根，我们得到第二个平方根：
( i 5 ) 1 2 = ( e i 5 π 2 ) 1 2 = e i 5 π 4 \Large (i^5)^{\frac{1}{2}} = (e^{i\frac{5\pi}{2}})^{\frac{1}{2}} = e^{i\frac{5\pi}{4}} (i5)21=(ei25π)21=ei45π
We can draw these two roots on an Argand diagram:
我们可以在复平面上画出这两个根：
We can do this again with i to the power 9 (which is also equal to i):
我们可以对
i
i
i 的
9
9
9 次幂（它也等于
i
i
i ）再次这样做：
( i 9 ) 1 2 = ( e i 9 π 2 ) 1 2 = e i 9 π 4 = e i ( π 4 + 2 π ) \Large (i^9)^{\frac{1}{2}} = (e^{i\frac{9\pi}{2}})^{\frac{1}{2}} = e^{i\frac{9\pi}{4}} = e^{i(\frac{\pi}{4}+2\pi)} (i9)21=(ei29π)21=ei49π=ei(4π+2π)
This gives a result that has an argument of
π
4
\frac{\pi}{4}
4π plus
2
π
2\pi
2π . Since adding
2
π
2\pi
2π has no effect on the value of a complex number, this result is identical to the original case where the argument was
π
4
\frac{\pi}{4}
4π . There are only two distinct square roots of i.
这给出一个结果，其辐角为
π
4
\frac{\pi}{4}
4π 加上
2
π
2\pi
2π 。由于加上
2
π
2\pi
2π 对复数的值没有影响，这个结果与辐角为
π
4
\frac{\pi}{4}
4π 的原始情况相同。
i
i
i 只有两个不同的平方根。
In fact, every complex number (except 0) has two distinct square roots, 3 distinct cube roots, and n distinct nth roots.
事实上，每个复数（除了
0
0
0 ）都有两个不同的平方根、三个不同的立方根和
n
n
n 个不同的
n
n
n 次根。
Takeaway 要点
i
i
i raised to a power p can sometimes have multiple values. Those values can be found by calculating the powers of the following equivalent numbers:
i
i
i 的
p
p
p 次幂有时可以有多个值。这些值可以通过计算以下等价数的幂来找到：
i p = e i p π 2 \Large i^p = e^{ip\frac{\pi}{2}} ip=eip2π
i p = e i p ( π 2 + 2 π ) \Large i^p = e^{ip(\frac{\pi}{2}+2\pi)} ip=eip(2π+2π)
i p = e i p ( π 2 + 4 π ) \Large i^p = e^{ip(\frac{\pi}{2}+4\pi)} ip=eip(2π+4π)
⋯
\Large\cdots
⋯
i
p
=
e
i
p
(
π
2
+
2
n
π
)
\Large i^p = e^{ip(\frac{\pi}{2}+2n\pi)}
ip=eip(2π+2nπ)
Not all of these roots are necessarily distinct.
并非所有这些根都一定是不同的。
i to the power i
So now we are in a position to calculate the value of
i
i
i to the power i. We will assume that we can raise i to the power
i
i
i simply by setting p to the value i in the formula above. This can be shown to be true, but we won’t prove it here.
Here is the result:
所以现在我们可以计算
i
i
i 的
i
i
i 次幂的值了。我们假设可以通过在上述公式中将
p
p
p 设为
i
i
i 的值来计算
i
i
i 的
i
i
i 次幂。这可以被证明是正确的，但我们这里不进行证明。
结果如下：
i i = e i ⋅ i π 2 = e − π 2 \Large i^i = e^{i\cdot i\frac{\pi}{2}} = e^{-\frac{\pi}{2}} ii=ei⋅i2π=e−2π
This is a very interesting result. The two i i i terms multiply to give -1, so the exponent is now a real number. This means that the power is a real number expression!
这是一个非常有趣的结果。两个
i
i
i 项相乘得到
−
1
-1
−1 ，所以指数现在是一个实数。这意味着幂是一个实数表达式！
i
i
i 的
i
i
i 次幂就是
−
π
2
-\frac{\pi}{2}
−2π 的指数形式。它的实数值约为
0.207880
0.207880
0.207880 。
But it gets a little weirder. We also have to consider the other possible results based on the alternate modulus-argument forms of i. For example when n equals 1, we add 2π to the exponent:
但事情变得更奇怪了。我们还必须考虑基于
i
i
i 的其他可能的模 - 辐角形式的结果。例如，当
n
=
1
n = 1
n=1 时，我们给指数加上
2
π
2\pi
2π ：
i = e i ( π 2 + 2 π ) = e i ⋅ 5 π 2 \Large i = e^{i(\frac{\pi}{2}+2\pi)} = e^{i\cdot\frac{5\pi}{2}} i=ei(2π+2π)=ei⋅25π
i i = e i ⋅ i 5 π 2 = e − 5 π 2 \Large i^i = e^{i\cdot i\frac{5\pi}{2}} = e^{-\frac{5\pi}{2}} ii=ei⋅i25π=e−25π
This gives a value of approximately 0.000388203.
这给出一个约为
0.000388203
0.000388203
0.000388203 的值。
We can use negative values of n too, of course. When n equals -1, we subtract 2π from the exponent:
当然，我们也可以使用
n
n
n 的负值。当
n
=
−
1
n = -1
n=−1 时，我们从指数中减去
2
π
2\pi
2π ：
i i = e i ( π 2 − 2 π ) = e − i 3 π 2 \Large i^i = e^{i(\frac{\pi}{2}-2\pi)} = e^{-i\frac{3\pi}{2}} ii=ei(2π−2π)=e−i23π
i i = e − i ⋅ i 3 π 2 = e 3 π 2 \Large i^i = e^{-i\cdot i\frac{3\pi}{2}} = e^{\frac{3\pi}{2}} ii=e−i⋅i23π=e23π
This gives a value of approximately 111.318.
这给出一个约为
111.318
111.318
111.318 的值。
Since this formula is based on the exponential function of a real number, every different value of n will give a unique, real result.
由于这个公式基于实数的指数函数，
n
n
n 的每个不同值都会给出一个唯一的实结果。
So i to the power i has an infinite number of solutions, and they are all real numbers.
所以
i
i
i 的
i
i
i 次幂有无限多个解，并且它们都是实数。
What Is I Raised To The Power I?
2023-10-19 00:00:00
What Is I Raised To The Power I?
An imaginary number raised to an imaginary number turns out to be real. However, while learning complex analysis, one learns that an exponential with respect to an imaginary number does not have a single, fixed value. Rather, the function is multi-valued — the value we arrived at in our calculation is just one of many values.
虚数的虚数次幂结果是实数。然而，在学习复分析时，人们会了解到关于虚数的指数函数并不具有单一、固定的值。确切地说，该函数是多值的——我们在计算中得到的值只是众多值中的一个。
The imaginary unit
i
i
i represents the square root of
−
1
-1
−1 , such that
i
2
=
−
1
i^2=-1
i2=−1 . Imaginary numbers live in a world of their own; the numbers are counted on an entirely different plane or axis that is solely devised for them. However, imaginary numbers have acquired a somewhat nefarious reputation, considering that their discovery has compounded the difficulty of problems that math was already replete with. I mean, as if the numbers we already had weren’t enough?
虚数单位
i
i
i 表示
−
1
-1
−1 的平方根，即
i
2
=
−
1
i^2 = -1
i2=−1 。虚数存在于它们自己的世界中；这些数是在一个完全不同的平面或轴上计数的，而这个平面或轴是专门为它们设计的。然而，虚数却获得了一个有点不好的名声，因为它们的发现增加了数学原本就充满的问题的难度。我的意思是，好像我们已有的数还不够多似的？
Our problem, however, combines not just one, but two of the many haunting aspects of mathematics. If
i
i
i itself is so difficult to comprehend, what could
i
i
i^i
ii be? You might be surprised to know that unlike
i
i
i , the value of
i
i
i^i
ii is a real number! How is that possible?
然而，我们的问题不仅结合了数学中众多令人困扰的方面中的一个，而是两个。如果
i
i
i 本身就如此难以理解，那么
i
i
i^i
ii 会是什么呢？你可能会惊讶地知道，与
i
i
i 不同，
i
i
i^i
ii 的值是一个实数！这怎么可能呢？
First, we must revise one of the most common, but least understood, mathematical operations — logarithms. Let me remind you how they work: If
a
=
b
c
a = b^c
a=bc , then
log
b
a
=
c
\log_b a = c
logba=c . For instance, given that
10
=
1
0
1
10 = 10^1
10=101 ,
log
10
10
=
1
\log_{10}10 = 1
log1010=1 . Here,
b
b
b is called the base of the logarithmic operation.
首先，我们必须复习一个最常见但却最不被理解的数学运算——对数。让我提醒你它们是如何工作的：如果
a
=
b
c
a = b^c
a=bc ，那么
log
b
a
=
c
\log_b a = c
logba=c 。例如，已知
10
=
1
0
1
10 = 10^1
10=101 ，则
log
10
10
=
1
\log_{10}10 = 1
log1010=1 。在这里，
b
b
b 被称为对数运算的底数。
Log scale (Photo Credit : Saim Tokacoglu/Shutterstock) 对数刻度尺
To solve the problem at hand, the base of our operation is
e
e
e , or Euler’s number, with a value of
2.71828
⋯
2.71828\cdots
2.71828⋯ .
为了解决手头的问题，我们运算的底数是
e
e
e ，即欧拉数，其值为
2.71828
⋯
2.71828\cdots
2.71828⋯。
If
e
y
=
x
e^y = x
ey=x , then
log
e
x
=
y
\log_e x = y
logex=y . The logarithm with the base
e
e
e is known as a “natural logarithm”. It is imperative to understand that the exponential and logarithmic functions are inverse functions. This is very important for our calculation.
如果
e
y
=
x
e^y = x
ey=x ，那么
log
e
x
=
y
\log_e x = y
logex=y 。以
e
e
e 为底的对数被称为“自然对数”。必须理解指数函数和对数函数是反函数。这对我们的计算非常重要。
log e e x = x ∵ log a b = b log a \Large \log_{e} e^{x} = x \quad \because \log a^{b} = b \log a logeex=x∵logab=bloga
e log e x = x ∵ log e x = y and e y = x \Large e^{\log_{e} x} = x \quad \because \log_{e} x = y \text{ and } e^{y} = x elogex=x∵logex=y and ey=x
The second thing to recall is that
i
i
i can be written as a complex number
z
=
0
+
i
z = 0 + i
z=0+i ， which can also be written as
z
=
cos
(
π
2
)
+
i
sin
(
π
2
)
z = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})
z=cos(2π)+isin(2π) 。 However, according to Euler’s formula,
e
i
x
=
cos
(
x
)
+
i
sin
(
x
)
e^{ix} = \cos(x) + i\sin(x)
eix=cos(x)+isin(x) 。 Therefore,
cos
(
π
2
)
+
i
sin
(
π
2
)
\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})
cos(2π)+isin(2π) is equal to
e
i
π
2
e^{i\frac{\pi}{2}}
ei2π 。
第二件要回忆的事情是
i
i
i 可以写成一个复数
z
=
0
+
i
z = 0 + i
z=0+i ，它也可以写成
z
=
cos
(
π
2
)
+
i
sin
(
π
2
)
z = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})
z=cos(2π)+isin(2π) 。然而，根据欧拉公式，
e
i
x
=
cos
(
x
)
+
i
sin
(
x
)
e^{ix} = \cos(x) + i\sin(x)
eix=cos(x)+isin(x) 。因此，
cos
(
π
2
)
+
i
sin
(
π
2
)
\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})
cos(2π)+isin(2π) 等于
e
i
π
2
e^{i\frac{\pi}{2}}
ei2π 。
i = e i π 2 i = e^{i\frac{\pi}{2}} i=ei2π
Now, let’s say
i
i
=
A
i^i = A
ii=A 。 Applying logarithm on both sides of the equation, we get:
现在，假设
i
i
=
A
i^i = A
ii=A 。在等式两边取对数，我们得到：
i i = A \Large i^i = A ii=A
log e i i = log e A \Large \log_e i^i = \log_e A logeii=logeA
i log e i = log e A \Large i\log_e i = \log_e A ilogei=logeA
i log e e i π 2 = log e A \Large i\log_e e^{i\frac{\pi}{2}} = \log_e A ilogeei2π=logeA
Remember that logarithmic and exponential functions are inverse functions, such that
log
e
e
x
=
x
\log_e e^x = x
logeex=x 。 Thus, from the above expression, one can discern:
记住对数函数和指数函数是反函数，即
log
e
e
x
=
x
\log_e e^x = x
logeex=x 。因此，从上面的表达式中，可以看出：
i ⋅ i π 2 = log e A \Large i\cdot i\frac{\pi}{2} = \log_e A i⋅i2π=logeA
i 2 π 2 = log e A \Large i^2\frac{\pi}{2} = \log_e A i22π=logeA
− π 2 = log e A \Large -\frac{\pi}{2} = \log_e A −2π=logeA
Applying the exponential on both sides of the equation:
在等式两边取指数：
e − π 2 = e log e A \Large e^{-\frac{\pi}{2}} = e^{\log_e A} e−2π=elogeA
Again, we encounter the inverse functions together. They neutralize to give the outcome:
再次，我们遇到了反函数在一起的情况。它们相互抵消得到结果：
e − π 2 = A \Large e^{-\frac{\pi}{2}} = A e−2π=A
i i = e − π 2 \Large i^i = e^{-\frac{\pi}{2}} ii=e−2π
i i ≈ 0.20788 \Large i^i\approx0.20788 ii≈0.20788
Thus, an imaginary number raised to an imaginary number turns out to be real. However, while learning complex analysis, one learns that an exponential with respect to
i
i
i does not have a single, fixed value. Rather, the function is multi-valued — the value we arrived at in our calculation is just one of many values. This is because the value of
i
i
i isn’t particularly
cos
(
π
2
)
+
i
sin
(
π
2
)
\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})
cos(2π)+isin(2π) ， but
cos
(
3
π
2
)
+
i
sin
(
3
π
2
)
\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})
cos(23π)+isin(23π) ，
cos
(
5
π
2
)
+
i
sin
(
5
π
2
)
\cos(\frac{5\pi}{2}) + i\sin(\frac{5\pi}{2})
cos(25π)+isin(25π) and every odd multiple of
π
2
\frac{\pi}{2}
2π 。 Therefore, the value of
i
i
i^i
ii also varies from
e
−
π
2
e^{-\frac{\pi}{2}}
e−2π to
e
−
3
π
2
e^{-\frac{3\pi}{2}}
e−23π ，
e
−
5
π
2
e^{-\frac{5\pi}{2}}
e−25π and so on.
因此，虚数的虚数次幂结果是实数。然而，在学习复分析时，人们会了解到关于
i
i
i 的指数函数并不具有单一、固定的值。确切地说，该函数是多值的——我们在计算中得到的值只是众多值中的一个。这是因为
i
i
i 的值不只是
cos
(
π
2
)
+
i
sin
(
π
2
)
\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})
cos(2π)+isin(2π) ，还有
cos
(
3
π
2
)
+
i
sin
(
3
π
2
)
\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})
cos(23π)+isin(23π) ，
cos
(
5
π
2
)
+
i
sin
(
5
π
2
)
\cos(\frac{5\pi}{2}) + i\sin(\frac{5\pi}{2})
cos(25π)+isin(25π) 以及
π
2
\frac{\pi}{2}
2π 的每一个奇数倍。因此，
i
i
i^i
ii 的值也从
e
−
π
2
e^{-\frac{\pi}{2}}
e−2π 变化到
e
−
3
π
2
e^{-\frac{3\pi}{2}}
e−23π ，
e
−
5
π
2
e^{-\frac{5\pi}{2}}
e−25π 等等。
图解虚数 i 的意义
作者：阮一峰 发表日期： 2012 年 9 月 24 日
有人在 Stack Exchange 问了一个问题：
" 我一直觉得虚数（imaginary number）很难懂。
中学老师说，虚数就是 - 1 的平方根
i = − 1 \Large i=\sqrt {-1} i=−1
可是，什么数的平方等于 - 1 呢？计算器直接显示出错！
直到现在，我也没有搞懂。谁能解释，虚数到底是什么？
它有什么用？"
帖子的下面，很多人给出了自己的解释，还推荐了一篇非常棒的文章
- A Visual, Intuitive Guide to Imaginary Numbers – BetterExplained
https://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/
我读后恍然大悟，醍醐灌顶，原来虚数这么简单，一点也不奇怪和难懂！
下面，我就用自己的语言，讲述所理解的虚数。
一、什么是虚数？
首先，假设有一根数轴，上面有两个反向的点：+1 和 - 1。
这根数轴的正向部分，可以绕原点旋转。显然，逆时针旋转 180 ° 180 ° 180°， + 1 +1 +1 就会变成 − 1 - 1 −1。
这相当于两次逆时针旋转 90 ° 90 ° 90° 。
因此，可以得到下面的关系式：
( + 1 ) ∗ ( 逆时针旋转 90 ° ) ∗ ( 逆时针旋转 90 ° ) = ( − 1 ) \large (+1) * (逆时针旋转 90 °) * (逆时针旋转 90 °) = (-1) (+1)∗(逆时针旋转90°)∗(逆时针旋转90°)=(−1)
如果把 + 1 + 1 +1 消去，这个式子就变为：
( 逆时针旋转 90 ° ) 2 = ( − 1 ) \large (逆时针旋转 90 °)^2 = (-1) (逆时针旋转90°)2=(−1)
将 " 逆时针旋转 90 ° 90 ° 90° " 记为 i i i ：
i 2 = ( − 1 ) \Large i^2 = (-1) i2=(−1)
这个式子很眼熟，它就是虚数的定义公式。
所以，我们可以知道，虚数 i i i 就是逆时针旋转 90 ° 90 ° 90°， i i i 不是一个数，而是一个旋转量。
二、复数的定义
既然
i
i
i 表示旋转量，我们就可以用
i
i
i ，表示任何实数的旋转状态。
将实数轴看作横轴，虚数轴看作纵轴，就构成了一个二维平面。旋转到某一个角°的任何正实数，必然唯一对应这个平面中的某个点。
只要确定横坐标和纵坐标，比如 ( 1 , i ) ( 1 , i ) (1,i)，就可以确定某个实数的旋转量 45 ° 45° 45°。
数学家用一种特殊的表示方法，表示这个二维坐标：用 + 号把横坐标和纵坐标连接起来。比如，把 ( 1 , i ) ( 1 , i ) (1,i) 表示成 1 + i 1 + i 1+i 。这种表示方法就叫做复数（complex number），其中 1 1 1 称为实数部， i i i 称为虚数部。
为什么要把二维坐标表示成这样呢，下一节告诉你原因。
三、虚数的加法
虚数的引入，大大方便了涉及到旋转的计算。
比如，物理学需要计算 “力的合成”。假定一个力是 3 + i 3 + i 3+i ，另一个力是 1 + 3 i 1 + 3i 1+3i ，请问它们的合成力是多少？
根据 “平行四边形法则”，你马上得到，合成力就是 ( 3 + i ) + ( 1 + 3 i ) = ( 4 + 4 i ) ( 3 + i ) + ( 1 + 3i ) = ( 4 + 4i ) (3+i)+(1+3i)=(4+4i)
这就是虚数加法的物理意义。
四、虚数的乘法
如果涉及到旋转角度的改变，处理起来更方便。
比如，一条船的航向是 3 + 4 i 3 + 4i 3+4i
如果该船的航向，逆时针增加 45 ° 45° 45°，请问新航向是多少？
45 ° 45° 45° 的航向就是 1 + i 1 + i 1+i 。计算新航向，只要把这两个航向 3 + 4 i 3 + 4i 3+4i 与 1 + i 1 + i 1+i 相乘就可以了（原因在下一节解释）：
( 3 + 4 i ) ∗ ( 1 + i ) = ( − 1 + 7 i ) \Large ( 3 + 4i ) * ( 1 + i ) = ( -1 + 7i ) (3+4i)∗(1+i)=(−1+7i)
所以，该船的新航向是 − 1 + 7 i -1 + 7i −1+7i 。
如果航向逆时针增加 90 ° 90° 90°，就更简单了。因为 90 ° 90° 90° 的航向就是 i i i ，所以新航向等于：
( 3 + 4 i ) ∗ i = ( − 4 + 3 i ) \Large ( 3 + 4i ) * i = ( -4 + 3i ) (3+4i)∗i=(−4+3i)
这就是虚数乘法的物理意义：改变旋转角度。
五、虚数乘法的数学证明
为什么一个复数改变旋转角度，只要做乘法就可以了？
下面就是它的数学证明，实际上很简单。
任何复数 a + b i a + bi a+bi，都可以改写成旋转半径 r r r 与横轴夹角 θ θ θ 的形式。
假定现有两个复数 a + b i a + bi a+bi 和 c + d i c + di c+di，可以将它们改写如下：
a + b i = r 1 ∗ ( c o s α + i s i n α ) \Large a + bi = r_{1} * ( cosα + isinα ) a+bi=r1∗(cosα+isinα)
c + d i = r 2 ∗ ( c o s β + i s i n β ) \Large c + di = r_{2} * ( cosβ + isinβ ) c+di=r2∗(cosβ+isinβ)
这两个复数相乘， ( a + b i ) ( c + d i ) ( a + bi )( c + di ) (a+bi)(c+di) 就相当于
r 1 ∗ r 2 ∗ ( c o s α + i s i n α ) ∗ ( c o s β + i s i n β ) \Large r_{1} * r_{2} * ( cosα + isinα ) * ( cosβ + isinβ ) r1∗r2∗(cosα+isinα)∗(cosβ+isinβ)
展开后面的乘式，得到
c o s α ∗ c o s β − s i n α ∗ s i n β + i ( c o s α ∗ s i n β + s i n α ∗ c o s β ) cosα * cosβ - sinα * sinβ + i ( cosα * sinβ + sinα * cosβ ) cosα∗cosβ−sinα∗sinβ+i(cosα∗sinβ+sinα∗cosβ)
根据三角函数公式，上面的式子就等于
c o s ( α + β ) + i s i n ( α + β ) \Large cos (α+β) + isin (α+β) cos(α+β)+isin(α+β)
所以，
( a + b i ) ( c + d i ) ＝ r 1 ∗ r 2 ∗ ( c o s ( α + β ) + i s i n ( α + β ) ) \Large ( a + bi )( c + di ) ＝ r_{1} * r_{2} * ( cos (α+β) + isin (α+β) ) (a+bi)(c+di) ＝ r1∗r2∗(cos(α+β)+isin(α+β))
这就证明了，两个复数相乘，就等于旋转半径相乘、旋转角度相加。
via:
-
GraphicMaths - i to the power i
https://graphicmaths.com/pure/complex-numbers/i-to-power-i/ -
I Power I: What Is I Raised To The Power I?
https://www.scienceabc.com/eyeopeners/what-is-i-raised-to-the-power-i.html -
i to the i is a Real Number – Math Fun Facts
https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/ -
虚数的意义 - 阮一峰的网络日志 作者： 阮一峰 日期： 2012 年 9 月 24 日
注：1、原文图片 csdn 转储失败，全部重传。2、全文公式 LaTex 化。
https://ruanyifeng.com/blog/2012/09/imaginary_number.html