# A conflicting result of Euler's formula

1. Jul 20, 2012

### Undoubtedly0

It can be shown from Euler's formula that

$$\left(-1\right)^x = \cos(\pi x) + i\sin(\pi x)$$

However, consider $x = 2/3$. The left expression gives

$$\left(-1\right)^\frac{2}{3} = \left(\left(-1\right)^2\right)^\frac{1}{3} = \left(1\right)^\frac{1}{3} = 1$$

while the right expression gives

$$\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$$

What has gone wrong here? Thanks all.

2. Jul 20, 2012

### andrien

because you have not done it in general.
(-1)^x=cos(n*pi*x)+isin(n*pi*x) where n is odd integer.now put x=2/3 and n=3.
you get it equal to 1.also check out that -1 has three cube roots so just check out that that complex number is arising because of other roots.

3. Jul 20, 2012

### DonAntonio

Complexpowers, roots don't behave as real ones. Read about multivaluate complex functions...or wait until the appropiate course at university.

DonAntonio

4. Jul 20, 2012

### Undoubtedly0

How does one derive $(-1)^x=\cos(n\pi x)+i\sin(n\pi x)$?

To derive $(-1)^x=\cos(\pi x)+i\sin(\pi x)$ I simply used
$$\ln\left((-1)^x)\right)=x\ln(-1)=x\pi i$$
therefore $(-1)^x = e^{x\pi i} = \cos(\pi x)+i\sin(\pi x)$.

Where did I miss the general case? Thanks again.

5. Jul 20, 2012

### andrien

can't you just verify that -1 is equal to exp(i*n*pi) when n is odd.

6. Jul 20, 2012

### Techian

The equation, (-1)^x=cos(pi*x)+i*sin(pi*x) is derived from Euler's identity based on the assumption that 'x' is an integer. Hence substituting fractional value for 'x' in the equation will lead to erroneous results. The derivation is as follows,
e^(i*y)=cos(y)+i*sin(y).....(Original Euler's identity)
Now replacing 'y' with 'pi*x', you will end up with
e^(i*pi*x)=cos(pi*x)+i*sin(pi*x)......(1)
Assuming that 'x' is an integer, cos(pi*x)=(-1)^x and sin(pi*x)=0.
substituting them in (1)
e^(i*pi*x)=(-1)^x.......(2)
substituting (2) in (1), you will end up with (-1)^x=cos(pi*x)+i*sin(pi*x) for x is an integer.