Solve Polar Coordinates: (-2*sqrt3) – (2*i) = 4*e^(i*7*pi/6)

[hugo28]

Question: Given (-2*sqrt3) – ( 2*i) = 4*e^(i*7*pi/6)

Perform the following:
(a) Convention I: angle go from (0) to 2*pi.
(b) Convention II: angle goes from (-)*pi to (+)*pi.
= = = = = = = = = = = = = = = = = = = = = = = = = = =

Part (a): Convention I: angle go from (0) to 2*pi.
r = (x^2 + y^2)^1/2 = sqrt16 = 4
(theta) = tan^(-1) |y/x =1/sqrt3| = pi/6
Since x = (-) and y = (-) it is in the third quadrant
(theta) = pi + pi/6 = 7*pi/6
Thus, z = r*e^(i*(theta)) = 4*e^(i*7*pi/6)


= = = = = = = = = = = = = = = = = = = = = = = = = = =
Part (b): Convention II: an angle goes from (-)*pi to (+)*pi.

Given convention II starts at (-)*pi to (+)*pi,
(theta) = tan^(-1) |y/x =1/sqrt3| = pi/6
Thus, the angle = pi/6 in the third quadrant becomes (-)*pi/6
Therefore: z = r*e^(i*(theta)) = 4*e^(i*7*pi/6) or
= 4*[cos(- pi/6) + i*sin(- pi/6)]
= 4*[sqrt3/2 – ½*i ]
= (2*sqrt3) – (2*i)

Result: is NOT the same as given (-2*sqrt3) – (2*i)


The convention II: an angle goes from (-)*pi to (+)*pi, that gives me most trouble.
At all possible would you please guide me through this (b)’s part and elucidate convention II, also. Sincerely.


Note: From my understanding and also can be wrong too that: from left hand side
at (-1)*pi rotates downward and toward right hand side to zero. From zero rotates upward and toward left hand side, is 0 to (+)*pi (counterclockwise direction).
Clockwise direction, from 0 to (-)*pi (like 3 to 9) is for z* (complex conjugate) only. Thanks.

 

[vela]

The convention is to measure angles from the +x axis, taking the counterclockwise direction to be the positive direction. So the angle $-\pi/6$ would actually be in the fourth quadrant, not the third. That's where you went wrong in part (b).