SOLUTION: Find the principal argument Arg z for the ff values of z. (-1-i√3)^2

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Question 1177967: Find the principal argument Arg z for the ff values of z.
(-1-i√3)^2
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
%28-1-i%2Asqrt%283%29%29%5E2
1%2B+2sqrt%283%29%2Ai%2Bi%5E2%2A3
1%2B+2sqrt%283%29%2Ai-3
-2+%2B+i+2sqrt%283%29+

=> z=a%2B+bi+++ In your case

a=-2+and b=2sqrt%283%29+=> a%3C0+ and b%3E0

To find argument we use the following formula:

theta=tan%5E-1%28b%2Fa%29+%2B180° if +a%3C0

theta=tan%5E-1%282sqrt%283%29%2F-2%29+%2B180°

theta=tan%5E-1%28-sqrt%283%29%29+%2B180°

theta=-60%2B180°

theta=120°

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The purely algebraic solution shown by the other tutor is fine, because the problem uses "nice" numbers.

deMoivre's Theorem gives us a tool for solving problems like this if the numbers aren't so nice.

(1) Represent the given number in polar form (I'll use degrees).

(-1,-i*sqrt(3)) = (2,240)

(2) deMoivre's Theorem says to square a complex number in polar form you square the magnitude and double the angle.

((-1,-i*sqrt(3))^2 = (2^2,2*240) = (4,480) = (4,120)

The problem only asked for the principal argument for the given expression, so you wouldn't be concerned with the magnitude. You would only have to know that the angle for (-1-i*sqrt(3)) is 240 degrees, so the angle for that number squared is 480 degrees, or 120 degrees.