Disproving Euler's Identity: Check My Math

[Onyxus]

I was messing around with Euler’s Identity and I think I accidently disproved it. I would like someone to check my math to make sure I didn’t make any rookie mistakes.

$$\begin{array}{l}<br /> e^{\pi i} + 1 = 0 \\ <br /> e^{\pi i} = - 1 \\ <br /> \left( {e^{\pi i} } \right)^2 = \left( { - 1} \right)^2 \\ <br /> e^{2\pi i} = 1 \\ <br /> \ln \left( {e^{2\pi i} } \right) = \ln \left( 1 \right) \\ <br /> 2\pi i = 0 \\ <br /> \frac{{2\pi i}}{{\pi i}} = \frac{0}{{\pi i}} \\ <br /> 2 = 0 \\ <br /> \end{array}$$

 

[micromass]

So, how did you define the logarithm on complex numbers??

 

[mathman]

ln(1) = 2kπi, where k is any integer. ln is a multivalued function.

 

[DonAntonio]

I was messing around with Euler’s Identity and I think I accidently disproved it. I would like someone to check my math to make sure I didn’t make any rookie mistakes.

$$\begin{array}{l}<br /> e^{\pi i} + 1 = 0 \\ <br /> e^{\pi i} = - 1 \\ <br /> \left( {e^{\pi i} } \right)^2 = \left( { - 1} \right)^2 \\ <br /> e^{2\pi i} = 1 \\ <br /> \ln \left( {e^{2\pi i} } \right) = \ln \left( 1 \right) \\ <br /> 2\pi i = 0 \\ <br /> \frac{{2\pi i}}{{\pi i}} = \frac{0}{{\pi i}} \\ <br /> 2 = 0 \\ <br /> \end{array}$$

What gives you away as a rookie is the title of your post, not your mathematics...which are also wrong.

Perhaps you'll be interested in reading about the complex logarithmic function's definition...

DonAntonio

 

[tony873004]

edit: student posted under my account

 

[Onyxus]

Thank you, DonAntonio, mathman and micromass, I thought that was what my error was, but I wasn't sure. You see, I haven't yet taken a course in which I learn even the basics of complex numbers, so my knowledge in that area is rather lacking.