In my homework I have given quite often this complex function to integrate on $$ \gamma(t)= e^{2i\pi t} $$ with $t\in[0,1]$ real. Now what I am asking myselfe is, isn't this $$ \gamma(t)= e^{2i\pi t}=(((e^i)^{\pi})^2)^t=((-1)^2)^t=1^t=1 $$ But this seems... very unlikely right?
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3The exponentiation rule $(a^b)^c=a^{bc}$ doesn't work in complex numbers. – Thomas Andrews Feb 05 '20 at 15:06
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Ah, thx Andrew! Of course. – KingDingeling Feb 05 '20 at 15:08
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It's a generalization of the problem that $$-1=(-1)^1=(-1)^{2/2}=\left((-1)^2\right)^{1/2}=1^{1/2}=1.$$ – Thomas Andrews Feb 05 '20 at 15:09
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Cf. this question – J. W. Tanner Feb 05 '20 at 15:14