Is there a way : 1i^(-2i), symbolic, ==> e^Pi ?
Jan 24, 2017
07:02 AM
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Jan 24, 2017
07:02 AM
Is there a way : 1i^(-2i), symbolic, ==> e^Pi ?
Hello, Everyone.
From :
Is there a way : 1i^(-2i), symbolic, ==> e^Pi ?
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Jan 24, 2017
07:44 AM
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Jan 24, 2017
07:44 AM
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Jan 24, 2017
07:44 AM
....Only so:
