...yes I really do hate them. A lot.
Does anyone know where to plot e^i?
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...yes I really do hate them. A lot.
Does anyone know where to plot e^i?
e^i is actually 1*e^(1*i) [r*e^iq polar form]
so,
r = Ö(a^2+b^2) = 1
a = 1.cosq
b = 1.sinq
q = 1 radian
=> a = cos1, b = sin1
Plot a,b on the argand diagram.
Run that past me again a bit slower will you?:o
what thinktank is saying is that e^i is in polar form
the standard form for a vector in polar form is like thinktank said...
r*e^(i*theta)
in your case, r, the coefficient is equal to 1, and so is theta
he is then solving for the values of a and b now that you have
r and theta where:
r = sqrt(a^2 + b^2)
a = r*sin (theta)
b = r*cos (theta)
again, r=1, theta = 1 rads
so a = 1*sin (1) and b = 1*cos(1)
figure out these values then plot them