(i)th Root and (i)th Power
Date: 02/13/99 at 11:01:58 From: Evan Murphy Subject: (i)th Root and (i)th Power What is the answer to x^i, if it is answerable at all? I asked my teacher, but he did not know. The related part of this question is the answer to x^(1/i). For example, is it possible to simplify 4^i 25^i 16^(1/i) 36^(1/i) ? I do not know how to answer these questions or even where to begin. Thanks. Evan Murphy
Date: 02/16/99 at 18:57:07 From: Doctor Luis Subject: Re: (i)th Root and (i)th Power Your question is very reasonable, and one should expect that there should be an answer. Indeed, the answer to your question is not as simple as you would think, but it IS possible to calculate the "i"-th root of a number, even though the answer is not unique - in fact, there are infinitely many such roots. What does it REALLY mean to raise a number to the 'i'-th power? Can the expression 2^i be given any meaning at all? For example, you know that x^2 is just shorthand for the expression x*x. That is, the exponent tells you how many factors of x are being multiplied by each other (for example, in x^4, or x*x*x*x, there are 4 factors). What about "imaginary" powers such as the "i"-th power, or even worse: what about complex powers like x^(2+3i)? Is it even meaningful to write imaginary or complex powers? The answer to these questions lies in the algebraic properties of exponents and logarithms, and also in a very interesting and useful property discovered long ago by a famous mathematician named Leonhard Euler. This result is as follows: e^(i*t) = cos(t)+i*sin(t) This result is often referred to as "Euler's Formula." The number "e" that you see on the left side of the equation is a very important constant in mathematics. It has the approximate value e = 2.718281828459045... and is also known as "Euler's constant." It is the base of natural logarithms, and it turns out that many formulas in mathematics have a very simple form when logarithms are taken with base 'e'. Of course, you have already met the number "i" that appears with the exponent on the left side of our equation. It is just sqrt(-1). The number "t" is just a real number that is being multiplied by the i. On the right side you see the complex number "cos(t)+i*sin(t)". Its real part is cos(t), which is a real number, and its imaginary part is sin(t), which is also a real number. If you have already studied trigonometry you will immediately recognize those two functions, which are called "cosine" and "sine" functions, respectively. (BEWARE - you must evaluate cos(t) and sin(t) in radian mode. Ask your teacher about this.) In summary, what Euler's formula is telling us is that if you take the real number e and raise it to an imaginary power (the "i*t" part of the formula), then the answer will be, in general, a complex number (the "cos(t)+i*sin(t)" part of the formula). Now we know the answer to the problem of raising the number to an imaginary power like "i"; "i" can also be written "i*1", which tells you that t = 1 (in radians). To show you how this works, let us go through an example. We will calculate e^(i*pi) ("pi" is the famous number 3.1415...) So, according to Euler's formula, we have: e^(i*pi) = cos(pi)+i*sin(pi) Now, cos(pi) happens to be -1, and sin(pi) is just 0 (you can get these with a calculator). So now we have: e^(i*pi) = (-1)+i*(0) = -1 The answer is -1, a rather unexpected result! But now you may be asking yourself... "Wait a minute, Euler's formula only holds when I am raising the number e to an imaginary power. What happens if I raise ANOTHER number, like 2, to an imaginary power, like 'i' ?" We are saved from this problem by all the properties of exponential and logarithmic functions. It turns out that you can write ANY exponential function in the form of "e raised to some power." Once you write your problem in this form, all you do is apply Euler's formula, since we DO know how to calculate the answer when the base of our exponent is e. Now, the exponential function y = e^x has the inverse y = ln(x) where ln is the natural logarithm. What it means for those two functions to be inverses of each other is the following: e^(ln(x)) = x and ln(e^x) = x Another property of logarithms that will prove useful for our purposes is: ln(a^x) = x*ln(a) Notice that if a is just the number e, the equation becomes ln(e^x) = x*ln(e) = x, with ln(e) = 1 (because e is the base of ln). Just as an example, let us calculate the value of 2^i. 2^i = e^(ln(2^i)) (from the first eqn above) = e^(i*ln(2)) (from the third eqn we just wrote) = cos(ln(2))+i*sin(ln(2)) (applying Euler's formula with t=ln(2) = 0.7692389 + i*0.6389612 (approximate value using calculator) So, as you can see, 2^i is a complex number whose real and imaginary parts are given above. What about 4^(i/3), for example? Applying the same method and reasoning as above you should conclude that the answer is cos(ln(4)/3) + i*sin(ln(4)/3). Or picking one of the problems you gave me, let us do 16^(1/i). Now, 1/i is just -i (this is because 1/i=i/i^2=i/-1=-i), so the problem is actually 16^(-i) = e^(ln(16^(-i))) (using e^(ln(x))=x ) = e^(-i*ln(16)) (using ln(a^x)=a*ln(x)) = cos(-ln(16))+i*sin(-ln(16)) (Euler's formula w/ t=-ln(16) So, using a calculator, you can see the answer is about -0.932687 - i*0.360686 So we can do imaginary powers. But what about complex powers? Can we really do e^(-1+5i)? Again, we are saved by the properties of exponents. In fact, for any numbers u and v, e^(u+v) = (e^u)*(e^v) This is just the familiar algebraic rule that "when you multiply two numbers expressed with the same base, you just raise the base to the sum of the exponents." Now, you can prove that this holds for complex numbers as well. In particular, e^(a + i*b) = e^a * e^(i*b) but i*b is an imaginary number. Therefore we know how to calculate e^(i*b) from Euler's formula. Moreover, we already know what e^a is because "a" is just a real number. So, using that simple property, we can figure out what e raised to a complex power will be. For example, let us calculate e^(-1+i*5) e^(-1+i*5) = e^(-1)*e^(i*5) (using the "addition rule") = e^(-1)*(cos(5)+i*sin(5)) (applying euler's formula to the factor e^(i*5) ) = 0.10435349-i*0.35276853 (approximate value. obtained using a calculator) There is a subtle point I must mention now. The answer we obtained above is just one of the infinitely many roots that exist. It turns out that the logarithm function is what is called a "multivalued" function. This means that it does not really give you a unique answer, but many answers. In fact, the ln(z) function gives you infinitely many answers. However, mathematicians usually only keep one of the solutions (called the principal solution), since if you know any one of the answers you can figure out immediately all the other solutions, even though there are infinitely many solutions. This is related to the problem of finding the square root of a number. As you may know, taking the square root of a positive number yields two distinct answers (one positive, one negative), either one of which is correct. The same thing happens with the logarithm function, only instead of just two distinct answers you get infinitely many distinct answers. (Forgive me if I do not go into detail at this point but it would take me a long time to explain it to you now.) Feel free to ask us again if you have any questions or if you are still curious about anything. And keep up the good work! - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
Date: 02/16/99 at 19:08:33 From: Doctor Schwa Subject: Re: (i)th Root and (i)th Power If you have the ith power, ith roots are no big deal. 1/i = -i, so 4^(1/i) = 4^(-i) = 1/4^i ... That is, ith roots are just the reciprocals of ith powers. Here is what we know offhand: (4^i)^i = 4^(i*i) = 4^(-1) = 1/4 should be true, but that is not much to go on. If I use some properties of logarithms, log(4^i) = i*log 4, and now I have a bit more of a clue. What kind of number should have a complex logarithm? Now, another way to look at it: e^x = (1 + x/n)^n for big n (compound interest formula), so I can find e^i by taking (1+i/n)^n and seeing what it approaches as n gets bigger and bigger. We also have: e^(a + bi) = e^a e^bi = e^a (e^b)^i Putting all this together, it turns out that the best thing to do is to define e^bi = cos b + i sin b (where b is an angle measured in radians) and thus 4^i = e^(i ln 4) = cos (ln 4) + i sin (ln 4). This is perhaps surprising and strange, but also beautiful and true. If you want to learn more about this, try searching the Dr. Math archives for DeMoivre or for "Euler's formula"; also see "About e" from the Dr. Math FAQ: http://mathforum.org/dr.math/faq/faq.e.html - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/
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