As the title mentions I am trying to prove that $i^2 = -1$ but am not sure about how to do so correctly. Any help would be very much appreciated.
Very sorry for the confusion, my teacher had asked to show that $i^2=1$ but I quickly realized this was wrong and corrected the question. i in this case is defined in polar form as $i = 1(cos(pi/2))+i(sin(pi/2))$
If you think complex number as ordered pair and isomorphic to R ^2 and form a field over `R. Defined by the additive operation as (a,b)+(c,d)= (,a+c,b+d) and multiplicative operation as (a,b)(c,d)= (ac-bd, ad+bc) now i ^2= (0,1)(0,1)=(-1,0) =-1.
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