Functions of Imaginary Numbers
Date: 7/31/96 at 4:14:18 From: Anonymous Subject: Functions of Imaginary Numbers I saw your answer about the problem of evaluating the value of i^i (i is imaginary number or Sqrt [1-]) There you used: ln (i^i) i^i = e i ln i = e i (Pi/2) i = e , since ln i = Pi/2 * i -(Pi/2) = e then we can obtain a real solution for the expression. However, does (ln i) itself exist? As far as I know, the argument of any logarithm has to be real positive number. If the expression (ln i) is valid then there must be a certain 'value' for (sin i) or (cos i) because it seems that imaginary number (i) can be treated as an ordinary real number. One last question.. is the equation iA e = cos A + i sin A a definition? if not, where does it come from? All these things are really confusing me... Thanks in advance!
Date: 9/1/96 at 18:10:19 From: Doctor Jerry Subject: Re: Functions of Imaginary Numbers Perhaps an analogy would be a good place to start. When students first start adding, they are taught to add the positive integers. The "add" function (add(x,y)=x+y), for young children, is defined for positive integers. Later on, the domain of the add function is extended to positive rational numbers, negative rationals, all real numbers, and to the set of all complex numbers. Much the same thing is true for the exponential function exp and the trigonometric functions sin and cos. One way of proceeding is through power series. The exponential function exp is defined for any complex number z = x + iy (x and y are real) by exp(z) = 1 + z/1! + z^2/2! + z^3/3! + ... It follows from this that exp(z) = exp(x+iy) = exp(x)*exp(iy) = exp(x)*(cos(y) + i*sin(y)). If you set x = 0 in this expression you will obtain exp(iy) = cos(y) + i*sin(y). Similarly, the sin and cos functions are defined for any complex number z = x + iy by power series. For example, sin(z) = z - z^3/3! + z^5/5! - ... If you just play around, not worrying about definitions too much, you might start with exp(iy) = cos(y) + i*sin(y) and replace y by -i. This gives exp(1) = cos(i) - i*sin(i). Similarly, replace y by i to get exp(-1) = cos(i) +i*sin(i). Now add to get exp(1)+exp(-1) = 2*cos(i). So, cos(i) = (exp(1)+exp(-1))/2 = cosh(1). Check this out on your calculator. HP48G calculates cos(i); I don't know about TI. Anyhow, cos(i) = (1.5430806...,0) = 1.5430806...+i*0 = 1.5430806... and cosh(1) = 1.5430806.... The course named complex variables is where all of this is explained. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 9/2/96 at 18:33:19 From: Doctor Pete Subject: Re: Functions of Imaginary Numbers Here I will address an alternate formulation of the principal value of i^i. Exp[I*t] = Cos[t]+I*Sin[t] For what value of t will the right-hand side of the above equal I? We note that for Cos[t] = 0, Sin[t] = 1, t = Pi/2. Hence Exp[I*Pi/2] = Cos[Pi/2]+I*Sin[Pi/2] = I . Raise both sides to the I(th) power. Thus I^I = (Exp[I*Pi/2])^I = Exp[I^2*Pi/2] = Exp[-Pi/2] . Alternatively, we may note that Log[z] = Log[Abs[z]] + I*(Arg[z]+2*Pi*k) , where k is any integer. Analytic continuation of the cosine and sine functions is also possible. To explain things a bit further, we need to discuss the nature of complex exponentiation and its inverse function. First, the "definition" Exp[I*t] = Cos[t]+I*Sin[t] is not so much a definition as it is a *natural* way of extending the exponentiation function to complex numbers. If you think of the complex plane, the real numbers is the x-axis in this plane. Thus, functions such as Sin[x] which were classically thought of as being "real" functions (i.e., having the reals as its domain), can often be "generalized" from the real line to the complex plane. This notion of generalization is more formally known as *analytic continuation.* This is not always possible, however; sometimes analytic continuation cannot be applied in a way such that the entire complex plane becomes a valid domain. There are many "proofs" that demonstrate the above identity, such as the Taylor series comparison. But ultimately these are more a result of more fundamental concepts than proofs in of themselves. Now, the interesting thing about complex exponentiation is that since it uses cosines and sines, it is a *many-to-one* mapping. More precisely, Sin and Cos are periodic functions, where Cos[x+2*Pi*k] = Cos[x] and Sin[x+2*Pi*k] = Sin[x] for any integer k. Therefore more than one complex number x+I*y will have the same exponential. This is important to keep in mind, because when one considers its inverse, the complex logarithm (as defined above), it becomes clear that it is a * one-to-many* mapping. That is, one complex logarithm has many (in fact, infinitely many) values. Which one you choose will depend on the situation. This is why there is some confusion when dealing with complex numbers- -their behavior is unusual because functions such as Exp and Log which were single-valued in the real domain suddenly become multivalued in the complex domain. Hope this helps--if you have any questions, please, feel free to ask. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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