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OpenStudy (anonymous):
Anyone else think that this is totally awesome?\[e^{i\pi}+1=0\]
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myininaya (myininaya):
\[e^{ix}=\cos(x)+i \sin(x) => e^{i \pi}=\cos(\pi)+i \sin(\pi)=-1+i (0)=-1\] yes its okay lol
OpenStudy (jamesj):
Totally
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