bainesy15
bainesy15
27-01-2016
Mathematics
contestada
what is e^(pi*i) using de moivre's?
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Аноним
Аноним
27-01-2016
you will Start with z=cos(x)+i*sin(x)......(1) , = i*z
so dz/z=i*dx , in (z) = i*x+c from (1); when x=0 | z=1 so c=0
in (z) = i*x
z=e^(ix) but z= cos (x)+i*sin
so cos (x)+i*sin (x) = e^ (i*x) (x) , Put x=pi in (z) & we get
ANSWER = -1+0=e^(i*p) , so it is = e^(i*pi)+1=0
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