This is how logarithms were originallycomputed by Mr. Briggs of Halifax, in 1620. He said, “I computedsuccessively 54 square roots of 10 .” We know he really computed only the first 27 , because the rest of them can be obtained by this trick with Δ . His work involved calculating the square root of 10 twenty-seven times, which is not much more than the ten times we did;however, it was more work because he calculated to sixteen decimal places, andthen reduced his answer to fourteen when he published it, so that there were norounding errors. He made tables of logarithms to fourteen decimal places bythis method, which is quite tedious. But all logarithm tables for three hundredyears were borrowed from Mr. Briggs’ tables by reducing the number ofdecimal places. Only in modern times, with the WPA and computing machines, havenew tables been independently computed. There are much more efficient methodsof computing logarithms today, using certain series expansions.
在1620年，哈利**斯的布里格斯先生，最初就是这样计算对数的。他说：“我计算了10的平方根，54个，连续的。”我们知道，他实际上只计算了27个，因为其他的，可以用Δ这个技巧，来得到。他的工作，包括计算10的平方根27次，与我们所做的10次相比，并不多很多；然而，他计算了16个小数位，这比我们多，然后，在公开发表时，他把的答案，缩减到14位，这样，就没有舍入误差了。通过这种方法，他创建了14位小数的对数表，这是非常冗长单调的工作。但是，随后的300年中，所有的对数表，都是从布里格斯先的表中，借来的，通过缩短小数位。只有在在现代，在WPA{？}和计算机器的帮助下，我们才有了新表，独立计算出来的。今天，计算对数，还有更高效的方法，即使用某些级数展开。

In the above process, we discoveredsomething rather interesting, and that is that for very small powers ϵwe can calculate 10ϵ easily; we have discovered that 10ϵ=1+2.3025ϵ, by sheer numerical analysis. Of course this also means that 10n/2.3025=1+nif n is very small. Now logarithms to any other base are merely multiplesof logarithms to the base 10 . The base 10 was used only because we have 10 fingers, and the arithmetic of it is easy, but if we ask for amathematically natural base, one that has nothing to do with the number offingers on human beings, we might try to change our scale of logarithms in someconvenient and natural manner, and the method which people have chosen is toredefine the logarithms by multiplying all the logarithms to the base 10 by 2.3025… This then corresponds to using some other base, and this is called thenatural base, or base e . Note that loge(1+n)≈n , or en≈1+n as n→0 .
在上面的过程中，我们发现，有些事情，相当有趣，那就是，对于非常小的次方ϵ，我们可以很容易地计算10ϵ；通过纯粹的数值分析，我们已经发现，10ϵ=1+2.3025ϵ 。当然，这也意味着，如果n非常小，那么，10n/2.3025=1+n 。现在，以任何数为底的对数，都只是以10为底的对数的复合。之所以，以10为底，乃是因为，我们有十个手指，其对数，比较容易，但是，如果我们要问，对于一个数学的自然底，一个与人类的手指数无关的底，我们或许可以尝试，以某种方便的和自然的方式，去改变我们对数的尺度，人们所选的方法，就是重新定义对数，让所有以10为底的对数，乘以2.3025…。因此，这就相当于，使用一个其他的底，这个底，被称为自然的底，或底e。注意，随着n→0，loge(1+n)≈n, 或 en≈1+n 。

It is easy enough to find out what eis: e=101/2.3025… or 100.434310… , an irrational power. Our table of the successive square roots of 10can be used to compute, not just logarithms, but also 10 to any power, so let us use it to calculate this natural base e. For convenience we transform 0.434310… into 444.73/1024 . Now, 444.73 is 256+128+32+16+8+4+0.73 . Therefore e , since it is an exponent of a sum, will be a product of the numbers
要找出e是什么，很容易，它就是：e=101/2.3025… 或100.434310… ,一个无理数的次方。我们的10的连续平方根的表，不仅可以用来计算对数，而且也可以计算10的任何次方，所以，让我们用它，来计算这个自然的底e。为了方便，我们把0.434310… 变成 444.73/1024。现在，444.73 就是256+128+32+16+8+4+0.73 。因此，由于e是一个和的指数，所以，它就是下面各数的乘积：
(1.77828)(1.33352)(1.074607)(1.036633)(1.018152)(1.009035)(1.001643)=2.7184.

(The only problem is the last one, which is 0.73 , and which is not in the table, but we know that if Δ is small enough, the answer is 1+0.0022486Δ .) When we multiply all these together, we get 2.7184 (it should be 2.7183 , but it is good enough). The use of such tables, then, is the way inwhich irrational powers and the logarithms of irrational numbers are allcalculated. That takes care of the irrationals.
（唯一的问题，就是最后一个，它是0.73，并不在表中，但是，我们知道，如果Δ足够小，答案就是1+0.0022486Δ。）当我们把所有这些数，都乘起来，我们就得到2.7184（它应该是2.7183，但也足够好了）。因此，对于无理数，其次方和对数，可以用这些表来计算。这种方式，可以处理无理数。

22–5Complex numbers 22-5 复数
Now it turns out that after all that workwe still cannot solve every equation! For instance, what is the squareroot of −1 ? Suppose we have to find x2=−1 . The square of no rational, of no irrational, of nothing thatwe have discovered so far, is equal to −1 . So we again have to generalize our numbers to a still wider class.Let us suppose that a specific solution of x2=−1 is called something, we shall call it i ; i has the property, by definition, that its square is −1 . That is about all we are going to say about it; of course, there ismore than one root of the equation x2=−1 . Someone could write i , but another could say, “No, I prefer −i . My i is minus your i .” It is just as good a solution, and since the only definitionthat i has is that i2=−1 , it must be true that any equation we can write is equally true if thesign of i is changed everywhere. This is called taking the complex conjugate.Now we are going to make up numbers by adding successive i ’s, and multiplying i ’s by numbers, and adding other numbers, and so on, according to allof our rules. In this way we find that numbers can all be written in the form p+iq, where p and q are what we call real numbers, i.e., the numbers we have beendefining up until now. The number i is called the unit imaginary number. Any real multiple of iis called pure imaginary. The most general number, a , is of the form p+iq and is called a complex number. Things do not get any worse if,for instance, we multiply two such numbers, let us say (r+is)(p+iq). Then, using the rules, we get
(22.4)
since ii= i2= −1 . Therefore all the numbers that now belong in the rules (22.1)have this mathematical form.
现在，在经过了这么多工作之后，结果还是，并非每个方程，我们都能解！例如，−1平方根，是多少？假设我们必须找出x2=−1。迄今为止，任何有理数、无理数、或任何数的平方，都不等于-1。所以，我们必须继续推广我们的数据，以获得更广泛的类型。假设x2=−1，有一个特殊的解，这个东西，我们将称其为i；通过定义，i有此属性，即其平方，等于-1。关于它，我们所要说的一切，就是这个；当然，对于方程x2=−1，有多于一个的根。某人可能会喜欢写i，但另外一人，可能会说：“不，我倾向于-i，我的i，是你的i的负数。”它同样是一个好的解，由于i所拥有的唯一定义，就是i2= −1，有一点，必须为真，即对于我们能写的任何方程，如果在任何地方，i的符号变了，但这些方程，应同样为真。这被称为：取复数共轭。现在，我们将依据我们所有的规则，通过增加相继的i、用数字乘以i、及增加其他的数等，来造数。以这种方式，我们发现，数可以被写作这种形式p+iq，这里p和q，都是实数，亦即，是我们迄今为止所定义的数。数字i，被称为单位虚数。任何实数乘以i，被称为纯虚数。最普遍的数，其形式为p+iq，被称为复数。如果我们把两个复数相乘，比如说(r+is)(p+iq)，则事情并不会变的更糟。因此，使用规则，我们就得到：
(22.4)
由于 ii= i2= −1 。因此，所有属于规则（22.1）的数，现在，就都具有这个数学形式。

Now you say, “This can go on forever! Wehave defined powers of imaginaries and all the rest, and when we are allfinished, somebody else will come along with another equation which cannot besolved, like x6+3x2=−2 . Then we have to generalize all over again!” But it turns out that withthis one more invention, just the square root of −1 , every algebraic equation can be solved! This is a fantasticfact, which we must leave to the Mathematics Department to prove. The proofsare very beautiful and very interesting, but certainly not self-evident. Infact, the most obvious supposition is that we are going to have to invent againand again and again. But the greatest miracle of all is that we do not. This isthe last invention. After this invention of complex numbers, we find that therules still work with complex numbers, and we are finished inventing newthings. We can find the complex power of any complex number, we can solve anyequation that is written algebraically, in terms of a finite number of thosesymbols. We do not find any new numbers. The square root of i , for instance, has a definite result, it is not something new; and iiis something. We will discuss that now.
现在，你说：“可以永远这样做”。我们已经定义了，虚数的次方、及其他东西，当我们做完之后，有人会带来其他的、无解的方程，比如x6+3x2=−2。然后，我们又需要重新推广数字，一切从头来过。”但结果则是，用这一个发明，即-1的平方根，所有的代数方程，都能解。这一事实，令人惊奇，我们应该把它，留给数学系去证明。证明是非常漂亮的和有趣的，但绝非不证自明。事实上，最明显的看法就是，我们将需一次、一次、又一次地去发明。但最大的奇迹则是，我们不需要。这就是最后的发明。在这个复数的发明出来之后，我们发现，规则对于复数，仍起作用。我们的发明工作，已经结束了。我们可以得出，任何复数的复数次方，任何方程，只要它是用代数写的，即是用有限的代数符号写的，我们都能解。我们不用去找任何新的数字。例如，i的平方根，有一个确定的结果，即它不是某个新的东西；ii是某个新的东西{？}。现在，我们将讨论之。

We have already discussed multiplication,and addition is also easy; if we add two complex numbers, (p+iq)+(r+is), the answer is (p+r)+i(q+s) . Now we can add and multiply complex numbers. But the real problem,of course, is to compute complex powers of complex numbers. It turns outthat the problem is actually no more difficult than computing complex powers ofreal numbers. So let us concentrate now on the problem of calculating 10 to a complex power, not just an irrational power, but 10(r+is). Of course, we must at all times use our rules (22.1)and (22.2).Thus
我们已经讨论了乘法，加法也很容易；如果我们把两个复数加起来，(p+iq)+(r+is) ,答案就是 (p+r)+i(q+s)。现在，对于复数，我们可以加和乘了。但是，真正的问题，则是计算：复数的复数次方。结果就是，这个问题，实际上，与计算实数的复数次方相比，并不更难。所以，现在，让我们集中注意，计算10的复数次方，这个问题，不是一个无理数的次方，而是10(r+is)。当然，我们应该一直使用我们的规则（22.1）和（22.2）。这样：
10(r+is)=10r10is (22.5)
But 10r we already know how to compute, and we can always multiply anything byanything else; therefore the problem is to compute only 10is . Let us call it some complex number, x+iy . Problem: given s , find x , find y . Now if
但是，我们已经知道，如何计算10r了，我们总是可以用某物乘以任何物；因此，这个问题，就只是计算10is 。让我们把它称为复数x+iy 。问题：给定s , 求x ,求y。现在如果：
10is=x+iy,
then the complex conjugate of this equation must also be true, so that
那么，这个方程的复数共轭，就应该也为真，于是：
10−is=x−iy.
(Thus we see that we can deduce a number of things without actuallycomputing anything, by using our rules.) We deduce another interesting thing bymultiplying these together:
（这样，我们看到，通过使用我们的规则，我们可以推导出一个东西的数字，而不用实际计算任何东西。）我们通过把这些东西乘在一起，推导出另外一个有趣的东西：
10is10−is=100=1=(x+iy)(x−iy)=x2+y2. (22.6)
Thus if we find x , we have y also.
这样，如果我们找到了x，那么，也就找到了y。

Now the problem is how to compute 10to an imaginary power. What guide is there? We may work over our rulesuntil we can go no further, but here is a reasonable guide: if we can computeit for any particular s , we can get it for all the rest. If we know 10is for any one s and then we want it for twice that s , we can square the number, and so on. But how can we find 10isfor even one special value of s ? To do so we shall make one additional assumption, which is not quitein the category of all the other rules, but which leads to reasonable resultsand permits us to make progress: when the power is small, we shall suppose thatthe “law” 10ϵ=1+2.3025ϵ is right, as ϵ gets very small, not only for real ϵ , but for complex ϵ as well. Therefore, we begin with the supposition that this law istrue in general, and that tells us that 10is=1+2.3025⋅is , for s→0 . So we assume that if s is very small, say one part in 1024 , we have a rather good approximation to 10is .
现在，问题就是，如何计算10的虚数次方。这里有什么指南呢？我们可以仔细研究我们的规则，直到我们不能走的更远，但是，这里是一个合理的指南：如果对于任何具体的s，我们可以计算它，那么，对于所有其他的数，我们就也可以计算。如果对于任何一个s，我们知道了10is，那么，如果我们想知道两倍s的虚数次方，我们就可以取该数的平方，如此等等。但是，即便是一个特殊的s的值，我们又如何才能找到其10is呢？要做到这点，我们将做一个附加的假设，它并不在所有规则的范畴之内，但是，它会导向一个合理的结果，并允许我们进步，它就是：当次方小的时候，我们将假设，“规律”10ϵ=1+2.3025ϵ是正确的，当ϵ变的非常小的时候，不仅对实数ϵ成立，对复数ϵ也成立。因此，我们假设，这个规律，普遍正确，我们就从此开始，且它告诉我们，对于 s→0，10is=1+2.3025⋅is。所以，我们认为，如果s非常小，比如说1/1024，那么，对于10is，我们就有了一个相当好的近似。

Now we make a table by which we can computeall the imaginary powers of 10 , that is, compute x and y . It is done as follows. The first power we start with is the 1/1024 power, which we presume is very nearly 1+2.3025i/1024 . Thus we start with
10i/1024=1.00000+0.0022486i, (22.7)
and if we keep multiplying the number by itself, we can get to ahigher imaginary power. In fact, we may just reverse the procedure we used inmaking our logarithm table, and calculate the square, 4 th power, 8 th power, etc., of (22.7),and thus build up the values shown in Table 22–3. Wenotice an interesting thing, that the x numbers are positive at first, but then swing negative. We shall lookinto that a little bit more in a moment. But first we may be curious to findfor what number s the real part of 10is is zero. The y -value would be 1 , and so we would have 10is=1i , or is=log10i . As an example of how to use this table, just as we calculated log102before, let us now use Table 22–3 tofind log10i .
现在，我们建了一张表，通过它，我们可以计算10的所有虚数次方，也就是说，计算x和y。做法如下。我们开始的第一个次方，是1/1024次方，我们假设它，非常接近于1+2.3025i/1024。这样，我们就从下式开始：
10i/1024=1.00000+0.0022486i, (22.7)
如果我们继续让这个数，乘以它自己，那么，我们就可以得到一个更高的虚数次方。在建立对数表时，我们有一个过程；事实上，我们只要反向这个过程，就可以了，然后，计算（22.7）的平方、四次方、八次方等，这样，就建立了表22-3中所显示的值。我们注意到一件有趣的事情，即x的数，开始时，是正的，但随后，变成了负的。稍后，我们将更仔细地查看这个事情。但是，首先，我们很好奇，对于什么样的s，10is的实数部分，才是零。y的值，应该是1。于是，我们就有10is=1i , 或is=log10i 。如何使用这个表，我们举个例子，就像我们前面计算log102那样，现在，让我们用表22-3，来找出log10i ：
Table 22–3 10i/1024 =1 + 0.0022486i 的连续平方

Which of the numbers in Table 22–3 do wehave to multiply together to get a pure imaginary result? After a little trialand error, we discover that to reduce x the most, it is best to multiply “512 ” by “128 .” This gives 0.13056+0.99159i . Then we discover that we should multiply this by a number whoseimaginary part is about equal to the size of the real part we are trying toremove. Thus we choose “64 ” whose y -value is 0.14349 , since that is closest to 0.13056 . This then gives −0.01308+1.00008i . Now we have overshot, and must divide by 0.99996+0.00900i. How do we do that? By changing the sign of i and multiplying by 0.99996−0.00900i (which works if x2+y2=1 ). Continuing in this way, we find that the entire power to which 10 must be raised to give i is i(512+128+64−4−2+0.20)/1024 , or 698.20i/1024 . If we raise 10 to that power, we can get i . Therefore log10 i=0.68184i .
表22-3中的哪个数，乘起来，才能得到一个纯虚数呢？在经过一番测试和错误之后，我们发现，要把x减到最小，最好是让“512 ” 乘以 “128 。” 这会给出0.13056+0.99159i。因此，我们发现，我们应该用一个数，乘以这个，此数的虚数部分，大约等于‘我们想要移动的那个数’的实数部分。这样，我们选择 “64 ” 其y值，是0.14349 , 接近于0.13056 。然后，这给出−0.01308+1.00008i。现在，我们做过头了，应该除以0.99996+0.00900i。我们是如何做到的呢？通过改变 i的符号，及乘以 0.99996−0.00900i (如果 x2+y2=1，这就成立)。以这种方式继续，我们发现，要提供i，那么，所求的10的次方，就应该是i(512+128+64−4−2+0.20)/1024 , 或者698.20i/1024 。如果我们求10的这个次方，我们就可以得到i。因此，log10 i=0.68184i。

22–6Imaginary exponents 22-6 虚指数
Table 22–4

To further investigate the subject of takingcomplex imaginary powers, let us look at the powers of 10 taking successive powers, not doubling the power each time, inorder to follow Table 22–3 furtherand to see what happens to those minus signs. This is shown in Table 22–4, inwhich we take 10i/8 , and just keep multiplying it. We see that x decreases, passes through zero, swings almost to −1 (if we could get in between p=10 and p=11 it would obviously swing to −1 ), and swings back. The y -value is going back and forth too.
取复数的次方，是一个主题，为了进一步研究它，让我们看取10的连续次方，不是每次都把次方加倍，为了遵循表22-3，并且要看看，负号发生了什么。结果见表22-4，在其中，我们取10i/8，只是不断乘它。我们看到，x在减少，通过零，几乎是向-1摆动（如果我们可以取p=10 和 p=11之间的值，它肯定会摆动到-1），然后，又往回摆动。y的值，也是来回摆动。

Figure 22–1 图22-1
In Fig. 22–1 the dotsrepresent the numbers that appear in Table 22–4, andthe lines are just drawn to help you visually. So we see that the numbers xand y oscillate; 10is repeats itself, it is a periodic thing, and as such, itis easy enough to explain, because if a certain power is i , then the fourth power of that would be i2 squared. It would be +1 again, and therefore, since 100.68i is equal to i , by taking the fourth power we discover that 102.72iis equal to +1 . Therefore, if we wanted 103.00i , for instance, we could write it as 102.72i times 100.28i . In other words, it has a period, it repeats. Of course, we recognizewhat the curves look like! They look like the sine and cosine, and we shallcall them, for a while, the algebraic sine and algebraic cosine. However,instead of using the base 10 , we shall put them into our natural base, which only changes thehorizontal scale; so we denote 2.3025s by t , and write 10is=eit , where t is a real number. Now eit=x+iy , and we shall write this as the algebraic cosine of t plus i times the algebraic sine of t . Thus
eit= + i . (22.8)
What are the properties of and ? First, we know, for instance, that x2+y2must be 1 ; we have proved that before, and it is just as true for base eas for base 10 . Therefore +=1 . We also know that, for small t , eit=1+it , and therefore is nearly 1 , and is nearly t , and so it goes, that all of the various properties of theseremarkable functions, which come from taking imaginary powers, are thesame as the sine and cosine of trigonometry.
在图22-1中，点代表了出现在表22-4中的数字，把它们连线，是为了直观。于是，我们看到，数字x和y，在震荡；10is代表着它自己，它是一个周期性的事物，由于本身如此，所以很容易解释，因为，如果{它的}某个次方是 i，那么，其4次方，就应该是i2的平方。它将又会是+1，从而，由于100.68i 等于 i ，通过取4次方，我们发现，102.72i等于 +1。因此，如果我们想知道，比如103.00i ，那么，我们就可以把它写成102.72i 乘以100.28i。换句话说，它有一个周期，它重复着。当然，我们可以认出，这个曲线，看上去像什么！它们像正弦和余弦函数，在一段时间内，我们将把它们称为代数正弦和代数余弦。然而，除了以10为底外，我们将把它们放到我们的自然底上，它只改变了水平尺度；所以，我们通过t，来标示2.3025s，且写 10is=eit , 这里t是一个实数。现在，eit=x+iy，我们将把此式写为：t的代数余弦，加上i乘以t的代数正弦。这样：
eit= + i . (22.8)
和 的属性，是什么呢？首先，比如我们知道，x2+y2应该是1；我们前面已经证明过这点，无论是以e为底，还是以10为底，都同样为真。因此+=1 。我们也知道，对于小的t，eit=1+it，因此， 接近于1 ，而 则接近于t，它们就是这样，即这些函数，值得注意，它们都来自于取虚数次方，它们的各种属性，与三角函数的正弦与余弦一样。

Is the period the same? Let us find out. eto what power is equal to i ? What is the logarithm of i to the base e ? We worked it out before, in the base 10 it was 0.68184i , but when we change our logarithmic scale to e , we have to multiply by 2.3025 , and if we do that it comes out 1.570 . So this will be called “algebraic π/2 .” But, we see, it differs from the regular π/2 by only one place in the last point, and that, of course, is theresult of errors in our arithmetic! So we have created two new functions in apurely algebraic manner, the cosine and the sine, which belong to algebra, andonly to algebra. We wake up at the end to discover the very functions that arenatural to geometry. So there is a connection, ultimately, between algebra andgeometry.
周期是相同的吗？让我们找出。 e的什么次方，等于I ? i的以e为底的对数，是什么？我们以前得到过它，以10为底时，它是0.68184i，但是，当我们把我们的对数尺度，换成e，我们就必须乘以2.3025，如果我们这么做了，则结果就是1.570。所以，这将被称为“代数的π/2”。但是，我们看到，它与规范的π/2的区别，仅在于最后的小数点，而这，当然，就是我们代数中的错误的结果！所以，我们在纯代数的方式中，创造了两个新的函数，余弦和正弦，它们应该属于代数，且只属于代数。最终，我们清醒了，我们所发现的函数，对于几何来说，正是自然的{?}。所以，在代数和几何之间，就有一个终极联系。

We summarize with this, the most remarkableformula in mathematics:
我们把这个，总结为数学中最著名的公式：
eiθ=cosθ+isinθ. (22.9)
This is our jewel.
这是我们的宝贝。

We may relate the geometry to the algebraby representing complex numbers in a plane; the horizontal position of a pointis x , the vertical position of a point is y (Fig. 22–2). Werepresent every complex number, x+iy . Then if the radial distance to this point is called r and the angle is called θ , the algebraic law is that x+iy is written in the form reiθ , where the geometrical relationships between x , y , r , and θ are as shown. This, then, is the unification of algebra and geometry.
我们可以通过在一个平面上，表示复数，来让几何与代数相关；一个点的水平位置，就是x，垂直位置，就是y（图22-2）。我们把每一个复数，表示为x+iy。因此，如果到这个点的半径，被称为r，角度是θ，那么，代数规律就是，x+iy 可被写为reiθ，这里，x , y , r , 和 θ之间的几何关系，如图所示。因此，这就是代数与几何的统一。

Fig. 22–2.x+iy=reiθ. 图22-2 .x+iy=reiθ