Express the complex number in trigonometric form. -3 + 3 square root of three i

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Express the complex number in trigonometric form. -3 + 3 square root of three i

Mathematics
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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 \[\Large a = -3\] \[\Large b = 3\sqrt{3}\]
yeah, that is the correct expression
quick question... what is the square root of i?

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

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