e^((i pi)/7) e^(pi i) = e^(i8pi/7) (just add the exponents).
So arg z = 8pi/7 + 2npi (n an integer)
Arg z is the value of arg z that is between -pi and pi. This is 8pi/7 - 2pi = -6pi/7
Hi,
Say you had a function of:
e^((i pi)/7) e^(pi i)
And you had to find Arg(z),
I'm getting a bit lost, i'm trying to convert it into x + iy form, so i can use tan^-1(y/x) to get the final answer,
so im using e^((8 pi i )/7) = cos(8pi/7) + i sin(8pi/7)
but I can't take it down any further, any help would be appreciated, thanks.
I'm very concerned here. You are asking about exercises using the polar form in the complex plane, yet you seem to have no clue what that means. The point (2, 2), representing 2+ 2i in the complex plane, has modulus (distance from (0, 0)) of and argument (angle with the positive real axis) : (2+ 2i)e^{i 5\pi/6)= 2\sqrt{2}e^{i \pi/4}e^{i 5\pi/6}= 2\sqrt{2}e^{i(\pi/4+ 5\pi/6}= 2\sqrt{2}e^{i(3 \pi/12+ 10\pi/12}= 2\sqrt{2}e^{13 \pi/12}[/tex].
Since the problem asks for the principal argument, there is no need at all for the " ". The principal argument is with n= 0.
Never, ever, convert problems like these to degrees!Let n = -1
So 5pi/6 - 2pi = -210deg?
i also expanded it all, getting tan^-1(x/y) = 5pi/12