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### (i)th Root and (i)th Power

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Date: 02/13/99 at 11:01:58
From: Evan Murphy
Subject: (i)th Root and (i)th Power

teacher, but he did not know. The related part of this question is the

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
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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
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
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)

-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
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
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.

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/
```
Associated Topics:
High School Exponents
High School Imaginary/Complex Numbers
High School Logs

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