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Find the argument of the complex number 
-2sqrt2+2sqrt2i in the interval 
0 <= theta < 2pi. Express your answer in terms of 
pi.
Answer:

Find the argument of the complex number 22+22i -2 \sqrt{2}+2 \sqrt{2} i in the interval 0 \leq \theta<2 \pi . Express your answer in terms of π \pi .\newlineAnswer:

Full solution

Q. Find the argument of the complex number 22+22i -2 \sqrt{2}+2 \sqrt{2} i in the interval 0θ<2π 0 \leq \theta<2 \pi . Express your answer in terms of π \pi .\newlineAnswer:
  1. Identify Quadrant: The complex number given is 22+22i-2\sqrt{2} + 2\sqrt{2}i. Here, a=22a = -2\sqrt{2} and b=22b = 2\sqrt{2}. Both aa and bb are negative and positive respectively, which places the complex number in the second quadrant.
  2. Calculate ba|\frac{b}{a}|: In the second quadrant, the argument θ\theta is πarctan(ba)\pi - \arctan(|\frac{b}{a}|). Let's calculate ba|\frac{b}{a}|:\newline|\frac{b}{a}| = |\frac{\(2\)\sqrt{\(2\)}}{\(-2\)\sqrt{\(2\)}}| = |\(-1| = 11
  3. Calculate arctan(1)\arctan(1): Now we calculate arctan(1)\arctan(1), which is known to be π4\frac{\pi}{4} or 4545 degrees.
  4. Calculate Argument Theta: Since the complex number is in the second quadrant, the argument theta is: θ=πarctan(1)=ππ4=3π4\theta = \pi - \arctan(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}

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