Did Google calculator make a mistake?

[Semjase]

Manipulating Euler's equation

I got this equation

 

(-1)^(1/pi)*e^i=1^(1/pi)

 

To the best of my knowledge this equation is correct.

 

If it is, why would Google calculator give a complex

value for for the left side of the equation when the

actual value should be 1.

 

[swansont]

e^i is real?

[Semjase]

e^i isn't real but the right side of the equation

has to be.

 

Is Euler's formula wrong?

 

Or could it be a problem with i

mathematics?

[swansont]

e^i isn't real but the right side of the equation

has to be.

 

Then how can you claim the equation is correct? The left side is complex.

 

Is Euler's formula wrong?

 

Or could it be a problem with i

mathematics?

 

Option 3

[Semjase]

I can't make that claim, I made a false

assumption that it should be correct

when breaking no algebraic rules deriving

the equation.

 

This is a significant problem mathematics.

[Cap'n Refsmmat]

How did you derive the equation?

[Semjase]

I derived the equation as follows

 

assuming e^(i*pi)=-1

 

(-1)*e^(i*pi)=1

 

((-1)*e^(i*pi))^(1/pi)=1^(1/pi)

 

which gives (-1)^(1/pi)*e^i=1

[Cap'n Refsmmat]

What you've shown is that the identity [math](a b)^x = a^x b^x[/math] does not hold when a and b have complex values. It also does not hold when they have negative values; for instance,

 

[math]1 = 1^{1/2} = (-1 \times -1)^{1/2} = (-1)^{1/2} (-1)^{1/2} = i \times i = -1[/math]

 

which is clearly false.

 

This is because exponentiation of a complex number is requires a choice of branch cut, since it is defined as [math]w^z = \exp(z \log w)[/math], and the logarithm of a complex number is a multi-valued function. (Similarly, the property fails for negative bases because the ordinary real definition of the logarithm doesn't extend to negative numbers.)

 

In short: the rules of math are just fine. You just broke one.

[Semjase]

I agree but since a and b are both real

a is negative and b is positive maybe

you could show me an example under

these conditions where

 

(a*b)^x is not equal to a^x*b^x

 

to settle any doubt in me for good.

Edited February 11, 2013 by Semjase

[John Cuthber]

There isn't any doubt.

You proved it yourself.

Your way led to a contradiction.

[Cap'n Refsmmat]

I agree but since a and b are both real

a is negative and b is positive maybe

you could show me an example under

these conditions where

 

(a*b)^x is not equal to a^x*b^x

 

to settle any doubt in me for good.

No, in your case you have both a and b negative, since [math]e^{i \pi} = -1[/math]. So my example applies.