anonymous
  • anonymous
Just need some homework help please. Find the polar for of the following complex number: sqrt3 - sqrt3i. I believe the answer is 3(cos pi/4 + i sin pi/4), am I correct?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1359476827595:dw|
anonymous
  • anonymous
angle should be either \(-\frac{\pi}{4}\) or \(\frac{7\pi}{4}\)
anonymous
  • anonymous
absolute value is \[\sqrt{\sqrt{3}^2+\sqrt{3}^2}\] \[=\sqrt{3+3}=\sqrt{6}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
the other answer options I have are 6(cos 7pi/4 + i sin 7pi/4) sqrt6(cos 7pi/4 - i sin 7pi/4) and sqrt6(cos 7pi/4 + i sin 7pi/4) I thought I worked it out correctly, guess not.
anonymous
  • anonymous
it is the last one
anonymous
  • anonymous
it helps to know what quadrant you are in so you can find the angle more easily \(\frac{\pi}{4}\) would put you in quadrant 1 but you are in quadrant 4
anonymous
  • anonymous
Yes, after seeing what you wrote I guess so. Thank you for that.
anonymous
  • anonymous
yw oh and don't forget \(|a+bi|=\sqrt{a^2+b^2}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.