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

Find the argument of the complex number 33i 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 33i 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}). Here, a=3a = 3 and b=3b = -\sqrt{3}.
  2. Use special triangles: Calculate the arctan of b/ab/a: θ=arctan(3/3)\theta = \arctan(-\sqrt{3}/3).
  3. Adjust for quadrant: Using the special triangles, we know that arctan(3/3)\arctan(-\sqrt{3}/3) corresponds to π/6-\pi/6 because tan(π/6)=3/3\tan(\pi/6) = \sqrt{3}/3, and since bb is negative, the angle is in the fourth quadrant.
  4. Find positive angle: However, the argument of a complex number is typically given as a positive angle between 00 and 2π2\pi. To find this, we add 2π2\pi to π/6-\pi/6 to get the positive angle.
  5. Calculate final argument: Adding 2π2\pi to π/6-\pi/6 gives us the final argument: θ=2ππ/6=(12π/6)(1π/6)=11π/6\theta = 2\pi - \pi/6 = (12\pi/6) - (1\pi/6) = 11\pi/6.

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