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Find the argument of the complex number 
9-3sqrt3i in the interval 
0 <= theta < 2pi.
Express your answer in terms of 
pi.
Answer:

Find the argument of the complex number 933i 9-3 \sqrt{3} i in the interval 0 \leq \theta<2 \pi .\newlineExpress your answer in terms of π \pi .\newlineAnswer:

Full solution

Q. Find the argument of the complex number 933i 9-3 \sqrt{3} i in the interval 0θ<2π 0 \leq \theta<2 \pi .\newlineExpress your answer in terms of π \pi .\newlineAnswer:
  1. Calculate arctan value: To find the argument of a complex number in the form a+bia + bi, where aa is the real part and bb is the imaginary part, we use the formula θ=arctan(ba)\theta = \text{arctan}(\frac{b}{a}). The complex number given is 933i9 - 3\sqrt{3}i, so a=9a = 9 and b=33b = -3\sqrt{3}.
  2. Simplify arctan expression: First, we calculate the arctan of b/ab/a, which is arctan(33/9)\text{arctan}(-3\sqrt{3}/9). Simplifying the fraction gives us arctan(3/3)\text{arctan}(-\sqrt{3}/3).
  3. Adjust for desired range: The value of arctan(3/3)\arctan(-\sqrt{3}/3) is known to be π/6-\pi/6 because tan(π/6)=3/3\tan(-\pi/6) = -\sqrt{3}/3. However, since we want the argument in the interval 0 \leq \theta < 2\pi, we need to add 2π2\pi to π/6-\pi/6 to find the equivalent angle in the desired range.
  4. Final argument calculation: Adding 2π2\pi to π/6-\pi/6 gives us 2ππ/62\pi - \pi/6, which simplifies to (12π/6)(π/6)=11π/6(12\pi/6) - (\pi/6) = 11\pi/6. Therefore, the argument of the complex number 933i9-3\sqrt{3}i in the interval 0 \leq \theta < 2\pi is 11π/611\pi/6.

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