I hate Argand diagrams...

[DavidHooper]

...yes I really do hate them. A lot.

Does anyone know where to plot e^i?

[thinktank2]

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.

[DavidHooper]

Run that past me again a bit slower will you?:o

[noble]

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