Why Does My Equation Suggest e^(πi) Equals Zero?

[jason17349]

if e^{\pi\imath}=-1 then:

-e^{\pi\imath}=1 and,

e^{2\pi\imath}=1

then:

-e^{\pi\imath}=e^{2\pi\imath}

\rightarrow e^{2\pi\imath}+e^{\pi\imath}=0

\rightarrow (e^{\pi\imath})^2+e^{\pi\imath}=0

\rightarrow (e^{\pi\imath}+1)e^{\pi\imath}=0

then:

e^{\pi\imath}=0

and

e^{\pi\imath}+1=0

Can somebody explain this contradiction to me?

 

[D H]

From the very first line you wrote:
e^{\pi i} + 1 = 0
You divided by zero when you went from
(e^{\pi i} + 1)e^{\pi i} = 0
to
e^{\pi i} = 0

 

[jason17349]

I thought if ab=0 then you could have two solutions a = 0 and b = 0?

 

[D H]

I thought if ab=0 then you could have two solutions a = 0 and b = 0?

Think again. For what values of x is 0\cdot x = 0 true?

 

[jason17349]

Whoops, sorry

 

[cristo]

I thought if ab=0 then you could have two solutions a = 0 and b = 0?

You're thinking of something like if ab=0, then either a=0 or b=0. However, in this case, you know that e^{\pi\imath}+1=0, and so (e^{\pi\imath}+1)e^{\pi\imath}=0 tells us nothing about e^{\pi i}