anonymous
  • anonymous
Express the complex number in trigonometric form. -3 + 3 square root of three i
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
I'm assuming the expression given is \(\Large -3+3\sqrt{3}*i\) If so, then it's the same as last time but now \[\Large a = -3\] \[\Large b = 3\sqrt{3}\]
anonymous
  • anonymous
yeah, that is the correct expression
Plasmataco
  • Plasmataco
quick question... what is the square root of i?

Looking for something else?

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

More answers

Plasmataco
  • Plasmataco
I know... im a little slow with the noggin but still...
jim_thompson5910
  • jim_thompson5910
the i isn't part of the square root
Plasmataco
  • Plasmataco
i know, just curios.
anonymous
  • anonymous
-1
Plasmataco
  • Plasmataco
...
Plasmataco
  • Plasmataco
i thought that was the square... exponent 2.
Plasmataco
  • Plasmataco
-1*-1=1 not i.
jim_thompson5910
  • jim_thompson5910
\[\Large i = \sqrt{-1}\] \[\Large i^2 = (\sqrt{-1})^2\] \[\Large i^2 = -1\]
anonymous
  • anonymous
\[\iota ^{2} = -1\]
Plasmataco
  • Plasmataco
am i just the stupid idiot here or u genuinely dont know or connection problemos?
Plasmataco
  • Plasmataco
square root. sry like radical.
jim_thompson5910
  • jim_thompson5910
sorry @Plasmataco I'm not following
Plasmataco
  • Plasmataco
i know but like -1 under a radical of 4
jim_thompson5910
  • jim_thompson5910
were you able to find r and theta, @lxoser ?
Plasmataco
  • Plasmataco
...
jim_thompson5910
  • jim_thompson5910
oh you mean \[\Large \sqrt{i} = \sqrt[4]{-1}\] @Plasmataco ??
Plasmataco
  • Plasmataco
\[\sqrt[4]{-1}\]
Plasmataco
  • Plasmataco
yeah
anonymous
  • anonymous
im gonna need help finding r and theta @jim_thompson5910
Plasmataco
  • Plasmataco
sry for interrupting...
anonymous
  • anonymous
its okay
jim_thompson5910
  • jim_thompson5910
\[\Large r = \sqrt{a^2 + b^2}\] \[\Large r = \sqrt{3^2 + (3\sqrt{3})^2}\] ... ... ... \[\Large r = ??\]
anonymous
  • anonymous
r = 30?
Plasmataco
  • Plasmataco
dont think so.
Plasmataco
  • Plasmataco
simplify.
Plasmataco
  • Plasmataco
it should be somthing a lot lower
jim_thompson5910
  • jim_thompson5910
Hint: \[\Large (3\sqrt{3})^2=(3\sqrt{3})*(3\sqrt{3})\] \[\Large (3\sqrt{3})^2=(3*3)*(\sqrt{3}*\sqrt{3})\] \[\Large (3\sqrt{3})^2=(3*3)*\sqrt{3*3}\] \[\Large (3\sqrt{3})^2=9\sqrt{9}\] \[\Large (3\sqrt{3})^2=9*3\] \[\Large (3\sqrt{3})^2=27\]
anonymous
  • anonymous
r = 6
jim_thompson5910
  • jim_thompson5910
yes
Plasmataco
  • Plasmataco
yup.
Plasmataco
  • Plasmataco
horay! now my question.
anonymous
  • anonymous
for theta, i got -60?
jim_thompson5910
  • jim_thompson5910
incorrect
anonymous
  • anonymous
is the result suppose to be in radians or degrees?
jim_thompson5910
  • jim_thompson5910
it depends on what the instructions say
jim_thompson5910
  • jim_thompson5910
does it say which mode they want?
anonymous
  • anonymous
no, but these are the answer choices
1 Attachment
jim_thompson5910
  • jim_thompson5910
ok so they want radian form
anonymous
  • anonymous
so would theta be - pi/3
jim_thompson5910
  • jim_thompson5910
yes, now because our point is in Q2 (see graph) we add pi radians to our angle to land in the right quadrant |dw:1442707471746:dw|
jim_thompson5910
  • jim_thompson5910
- pi/3 is in Q4 |dw:1442707591362:dw|
jim_thompson5910
  • jim_thompson5910
so we have to add 180 degrees or pi radians to get it into the right quadrant |dw:1442707637525:dw|
anonymous
  • anonymous
2pi/3?
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
correct
anonymous
  • anonymous
so would the answer be 6(cos 2pi/3 + i sin 2pi/3) ?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
thank you so much!! :")
jim_thompson5910
  • jim_thompson5910
no problem

Looking for something else?

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