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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/   
    
Associated Topics:
College Imaginary/Complex Numbers
High School Functions
High School Imaginary/Complex Numbers

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