**Question Corner and Discussion Area**

How do you defineYou can find a detailed answer to this question by reading the answers to some of the other questions on this site, but here is a summary of the answer all together in one place.a^(b+ci)?a,b,care real numbers.example: 5^(3+2

i)=?Thanks

The ordinary definition of exponentiation of real numbers
(*a*^*x*) only makes sense when *x* is rational. To extend the definition
to irrational and then to complex values of *x*, you need to rewrite
the definition in a way that makes sense even when *r* is complex.

One way to do this is to use the fact that *e*^*x* can be expressed as
the infinite sum

(where *n*! means *n* factorial, the product of the numbers
1,2,. . . ,*n*).

It makes perfectly good sense to add and multiply complex numbers, and
the theory about infinite sums can also be extended to complex numbers,
so this formula can be used as a definition of what *e*^*x* means when
*x* is complex.

If *x* is a "purely imaginary" number, that is, if *x*=*ci* where *c* is
real, the sum is very easy to evaluate, using the fact that *i*^2=-1,
*i*^3=-*i*, *i*^4=1, *i*^5=*i*, etc. When you do this and split the sum into
its real and imaginary parts, you find that the real part is the same
as the infinite sum expression for cos *c*, and the imaginary part is
the same as the infinite sum expression for sin *c*. This gives rise
to *de Moivre's formula*:

This is explained in more detail in the answer to another question. Also given in that answer is an alternative explanation which doesn't rely on the fact that functions likee^(ic) = (cosc) +i(sinc)

Now we know what *e* raised to an imaginary power is. One can also show that
the definition of *e*^*x* for complex numbers *x* still satisfies the usual
properties of exponents, so we can find *e* to the power of any complex number
*b *+ *ic* as follows:

Finally, for a real numbere^(b+ic) = (e^b)(e^(ic)) = (e^b)((cosc) +i(sinc))

This answers the question you asked. Now, if *a* is a complex number
instead of a real number, things are more complicated. There is no single
value to "ln *a*": there are lots of different complex numbers
*z* for which *e*^*z *= *a*, and for any such complex number *z*, you could
define *a*^(*b*+*ic*) to be *e*^(*z*(*b*+*ic*)) and use the above technique
to calculate it. This is illustrated in the
answer to another question, where it is shown that
the expression *i*^*i* has an infinite set of possible values.

In fact, the same thing is true even when *a* is a real number. Technically
speaking, the expression *a*^(*b*+*ic*) has infinitely many possible values
(except when *b* and *c* are both rational numbers), because instead
of doing the calculation writing *a *= *e*^(ln *a*), you could also
do it by writing , or by writing
, or ,
and so on. Each of these equalities is true (you can check them using
de Moivre's formula to show that , so adding multiples
of to the exponent is the same as multiplying by 1, which doesn't
affect the truth of the equality). Using these different equalities
to do the calculation gives rise to infinitely many different possible
values for *a*^(*b*+*ci*).

When *a* is real it is more "natural" to use the ordinary
real-valued logarithm ln *a* rather than than something
like . So it is reasonable to think of
*a*^(*b*+*ic*) as having only one value (in much the same way as we think
of 4^(1/2) as equalling 2 even though both 2 and -2 are square
roots of 4). Technically, this value is called the *principal value*.
This is what the formula up above gives you.

However, when *a* is not real there is no one natural choice of logarithm
to prefer over any other, so in those cases we have to say that an expression
like *a*^(*b*+*ic*) has many different values.

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Last updated: April 19, 1999

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