       When is x^y = != E^(y*Log[x])

• To: mathgroup at smc.vnet.net
• Subject: [mg66236] When is x^y = != E^(y*Log[x])
• From: ted.ersek at tqci.net
• Date: Fri, 5 May 2006 05:02:22 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```The Mathematica documentation for Power says:
For complex numbers (x^y) gives the principal value of  ( E^(y*Log[x] ).
This is consistent with reference books.

I wanted to see where in the extended complex plane this identity applies.
Also when it doesn't apply how do we determine (x^y).
So consider the following:

In:=
x=0; y=-3;
{x^y, E^(y*Log[x])}

Out=
{ComplexInfinity, Infinity}

The documentation wasn't wrong in the above example because 0, -3 are not
complex numbers.
However, I have seen books that imply the identity above works for any (x,y).
Well I can see the above identity doesn't apply when x=0 and y is negative.
The above identity doesn't apply in the following cases either.

In:=
x=-2;  y=(2+I)*Infinity;
{x^y, E^(y*Log[x])}

Out=
{Indeterminate, 0}

In:=
x=5-6*I;  y=-Infinity*I;
{x^y, E^(y*Log[x])}

Out=
{Indeterminate, 0}

In:=
x=y=Infinity;
{x^y, E^(y*Log[x])}

Out=
{ComplexInfinity, Infinity}

Then you might say the above identity doesn't apply when Abs[y]==Infinity,
but the above identity does apply in the next example.

In:=
x=(1+2*I)*Infinity;  y=(-1+2*I)*Infinity;
{x^y, E^(y*Log[x])}

Out=
{0, 0}

Could somebody provide conditions on (x,y) that are necessary and
sufficient for
E^(y*Log[x])  to return the same thing as (x^y).

-------------
Ted Ersek

```

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