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Functions of Imaginary Numbers

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Date: 7/31/96 at 4:14:18
From: Anonymous
Subject: Functions of Imaginary Numbers

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...

```

```
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
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
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).
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
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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