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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 9+33i 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 9+33i 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. Use formula for argument: 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 = \arctan(\frac{b}{a}). Here, a=9a = 9 and b=33b = 3\sqrt{3}.
  2. Calculate arctan function: Calculate the argument using the arctan function: θ=arctan(339)\theta = \arctan(\frac{3\sqrt{3}}{9}).
  3. Simplify fraction: Simplify the fraction inside the arctan function: θ=arctan(3/3)\theta = \arctan(\sqrt{3}/3).
  4. Recognize angle in triangle: Recognize that arctan(3/3)\arctan(\sqrt{3}/3) corresponds to the angle whose tangent is 3/3\sqrt{3}/3. This is a well-known angle in a 3030-6060-9090 right triangle, where the angle opposite the side with length 3\sqrt{3} is 6060 degrees or π/3\pi/3 radians.
  5. Determine final argument: Since the complex number is in the first quadrant (both aa and bb are positive), the argument is directly the arctan\arctan value: θ=π3\theta = \frac{\pi}{3}.

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