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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 3+3i 3+\sqrt{3} i in the interval 0θ<2π 0 \leq \theta<2 \pi .\newlineExpress your answer in terms of π \pi .\newlineAnswer:

Full solution

Q. Find the argument of the complex number 3+3i 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 Angle θ\theta: To find the argument of the complex number 3+3i3+\sqrt{3}i, we need to calculate the angle θ\theta that the line connecting the origin to the point (3,3)(3, \sqrt{3}) makes with the positive x-axis in the complex plane. The argument is given by θ=arctan(imaginary partreal part)\theta = \arctan(\frac{\text{imaginary part}}{\text{real part}}).
  2. Find Arctan: Calculate the arctan\arctan of the imaginary part divided by the real part: arctan(3/3)\arctan(\sqrt{3}/3).
  3. Identify Triangle Angle: 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.
  4. Determine Final Argument: Since the complex number 3+3i3+\sqrt{3}i is in the first quadrant (both real and imaginary parts are positive), the argument θ\theta is simply π/3\pi/3.

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