分栏双语对照

This is the most important function in mathematics. It is defined, for every complex number z, by the formula

在数学中这是一个最重要的函数。它是用公式来定义的，对每个复数z，规定

The series ⦅1⦆ converges absolutely, for every z and converges uniformly on every bounded subset of thc complex plane. Thus exp is a continuous function. The absolute convergence of ⦅1⦆ shows that the computation

级教⦅1⦆对每个z绝对收敛，对复平面的每个有界子集一致收敛。因此，exp是连续函数。⦅1⦆的绝对收敛指出了算式

\sum_{k = 0}^{∞} \frac{a^{k}}{k!} \sum_{m = 0}^{∞} \frac{b^{m}}{m!} = \sum_{n = 0}^{∞} \frac{1}{n!} \sum_{k = 0}^{n} \frac{n!}{k! (n - k)!} a^{k} b^{n - k} = \sum_{n = 0}^{∞} \frac{(a + b)^{n}}{n!}

is correct. It gives the important addition formula

是正确的。它给出了重要的加法公式

valid for all complex numbers a and b.

此公式对所有复数a和b是正确的。

We define the number e to be exp(1), and shall usually replace exp(z) by the customary shorter expression e^x. Note that e⁰ = exp(0) = 1, by ⦅1⦆.

我们规定数e是exp(1)。习惯上常用较短的表达式e^x代替exp(z)。注意，由⦅1⦆可得e⁰ = exp(0) = 1。

Theorem

For every complex z we have e^x ≠ 0.
exp is its own derivative: exp′(z) = exp(z).
The restriction of exp to the real axis is a monotonically increasing positive function, and e^x→∞ as x→∞, e^x→0 as x→-∞.
There exists a positive number π such that e^πi/2 = i and such that e^z = 1 if and only if z/(2πi) is an integer.
exp is a periodic function, with period 2πi.
The mapping t→e^it maps the real axis onto the unit circle.
If w is a complex number and w ≠ 0, then w = e^z for some z.

定理

对每一个复数z，e^x ≠ 0。
exp的导数是它自己：exp′(z) = exp(z)。
exp限制在实轴上是单调增加的正函数，且当x→∞时e^x→∞；当x→-∞时e^x→0。
存在一个正数π使得e^πi/2 = i，并使得e^z = 1当且仅当z/(2πi)是整数，
exp是周期函数，其周期是2πi。
映射t→e^it将实轴映到单位圆上。
若w是复教且w ≠ 0，则存在某个z使w = e^z。

By ⦅2⦆, e^z ⋅ e^-z = e^z-z = e⁰ = 1. This implies ⦅a⦆. Next,

由⦅2⦆，e^z ⋅ e^-z = e^z-z = e⁰ = 1。由此得到⦅a⦆。其次，

The first of the above equalities is a matter of definition, the second follows from ⦅2⦆, and the third from ⦅1⦆, and ⦅b⦆ is proved.

在上述等式中．第一个是定义，第二个从⦅2⦆得到，而第三个从⦅1⦆得到，因此证明了⦅b⦆。

That exp is monotonically increasing on the positive real axis, and that e^x→∞ as x→∞, is clear from ⦅1⦆. The other assertions of ⦅c⦆ are consequences of e^x ⋅ e^-x = 1。

由于⦅1⦆。显然exp在正实轴上是单调增加的．而且当x→∞时e^x→∞。⦅c⦆的另一个断言是e^x ⋅ e^-x = 1的推论。

For any real number t, ⦅1⦆ Shows that e^-it is the complex conjugate Of e^it。Thus

对于任何实数t，⦅1⦆表示e^-it是e^it的共轭复数。因此

In other words, if t is real, e^it lies on the unit circle. We define cos t, sin t to be the real and imaginary parts of e^it:

换句话说，若t为实数．则e^it位于单位圆上。我们定义cos t，sin t为e^it的实部和虚部：

If we differentiate both sides of Euler′s identity

若对等价于⦅4⦆的欧拉恒等式

which is equivaIent to ⦅4⦆, and if we apply ⦅b⦆, we obtain

两边微分，并且应用⦅b⦆，则得

The power series ⦅1⦆ yields the representation

幂级数⦅1⦆给出表示式

Take t = 2. The terms of the series ⦅7⦆ then decrease in absolute value (except for the first one) and their signs alternate. Hence cos 2 is less than the sum of the first three terms of ⦅7⦆, with t = 2; thus cos 2 < -⅓. Since cos 0 = 1 and cos is a continuous real function on the real axis, we conclude that there is a smallest positive number t₀ for which cos t₀ = 0. We define

取t = 2，则级数⦅7⦆的各项按绝对值减少（除首项外），而且它们的符号是交错的．因此cos 2小于级数⦅7⦆的前三项之和；于是cos 2 < -⅓。由于cos 0 = 1且cos是实轴上的实连续函数．故可断定存在一个最小的正数t₀使得cos t₀ = 0。我们定义

It follows from ⦅3⦆ and ⦅5⦆ that sin t₀ = ± 1. Since

从⦅3⦆及⦅5⦆得到sin t₀ = ± 1。由于在开区间(0, t₀)上，有

on the segment (0, t₀) and since sin 0 = 0, we have sin t₀ > 0, hence sin t₀ = 1, and therefore

又sin 0 = 0，故有sin t₀ > 0，因此sin t₀ = 1，而且

It follows that e^πi = i² = -1, e^2πi = (-1)² = 1, and then e^2πin = 1 for every integer n. Also, ⦅e⦆ follows immediately:

由此可见，e^πi = i² = -1，e^2πi = (-1)² = 1，并且对每个正整数n，e^2πin = 1。同样立即得到⦅e⦆:

If z = x + iy, x and y real, then e^z = e^xe^iy; hence |e^z| = e^x. If e^z = 1, we therefore must have e^x = 1, so that x = 0; to prove that y/(2π) must be an integer, it is enough to show that e^iy ≠ 1 if 0 < y < 2π, by ⦅10⦆.

若z = x + iy，x和y为实教，则e^z = e^xe^iy；因此|e^z| = e^x。若e^z = 1，则必须有e^x = 1，从而x = 0；根据⦅10⦆，为了证明y/(2π)一定是整数，只要证明当0 < y < 2π时，e^iy ≠ 1就足够了。

Suppose 0 < y < 2π, and

设0 < y < 2π，且

Since 0 < y/4 < π/2, we have u > 0 and v > 0. Also

由于0 < y/4 < π/2，故有u > 0，v > 0。同样

The right side of ⦅12⦆ is real only if u² = v²; since u² + v² = 1, this happens only when u² = v² = ½, and then ⦅12⦆ shows that

仅当u² = v²时，⦅12⦆的右边才是实数；由于u² + v² = 1仅当u² = v² = ½时才成立，因此⦅12⦆表明

This completes the proof of ⦅d⦆.

这就证明了⦅d⦆。

We already know that t → e^it maps the real axis into the unit circle. To prove ⦅f⦆, fix w so that |w| = 1; we shall show that w = e^it for some real t. Write w = u + iv, u and v real, and suppose first that u ≧ 0 and v ≧ 0. Since u ≦ 1, the definition of π shows that there exists a t, 0 ≦ t ≦ π/2, such that cos t = u; then sin² t = 1 - u² = v², and since sin t ≧ 0 if 0 ≦ t ≦ π/2, we have sin t = v. Thus w = e^it.

我们已经知道，t → e^it将实轴映入单位圆内。为了证明⦅f⦆，现固定w使得|w| = 1。我们将要证明，存在某个实数t使w = e^it。记w = u + iv，u和v为实数．而且首先假定u ≧ 0及v ≧ 0。由于u ≦ 1，则π的定义表明存在一个t，0 ≦ t ≦ π/2，使得cos t = u；因而sin² t = 1 - u² = v²，又由于当0 ≦ t ≦ π/2时有sin t ≧ 0，故sin t = v。因此w = e^it。

If u < 0 and v ≧ 0, the preceding conditions are satisfied by -iw. Hence -iw = e^it for some real t, and w = e^i(t+π/2). Finally, if v < 0. the preceding two cases show that -w = e^it for some real t, hence w = e^i(t+π) This completes the proof of ⦅f⦆.

若u < 0而v ≧ 0，则-iw满足上述条件。因此，存在某个实数t使-iw = e^it，而且w = e^i(t+π/2)。最后，若v < 0，则上述两种情况证明了，存在某些实数t使-w = e^it，因此w = e^i(t+π)。这就证明了⦅f⦆。

If w ≠ 0, put α = w / |w|. Then w = |w|α. By ⦅c⦆, there is a real x such that |w| = e^x. Since |α| = 1, ⦅f⦆ shows that α = e^iy for some real y. Hence w = e^{x + iy}. This proves ⦅g⦆ and completes the theorem. ▮

若w ≠ 0令α = w / |w|，因而w = |w|α。根据⦅c⦆，有一个实数x使得|w| = e^x。由于|α| = 1，则⦅f⦆证明了，对于某些实数y有α = e^iy。因此w = e^{x + iy}。这就证明了⦅g⦆，并且完成了定理的证明。 ▮

We shall encounter the integral of (1 + x²)^-1 over the real line. To evaluate, put φ(t) = sin t / cos t in (-π/2, π/2). By ⦅6⦆, φ′ = 1 + φ². Hence φ is a monotonically increasing mapping of (-π/2, π/2) onto (-∞, ∞), and we obtain

我们将遇到(1 + x²)^-1在实线上的积分。为了求它的值，在(-π/2, π/2)内，令φ(t) = sin t / cos t ，根据⦅6⦆有φ′ = 1 + φ²。因此，φ是一个(-π/2, π/2)到(-∞, ∞)上的单调增加的映射，而且得到

Real and Complex Analysis

实分析与复分析

Prologue

The Exponential Function

引言

指数函数

Theorem

定理

Proof

证明

Comment

原注

译注