Question

Express the complex number in trigonometric form.

-3 + 3 square root of three i

Answer 1

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

\[\Large a = -3\]
\[\Large b = 3\sqrt{3}\]

Answer 2

yeah, that is the correct expression

Answer 3

quick question... what is the square root of i?

Answer 4

I know... im a little slow with the noggin but still...

Answer 5

the i isn't part of the square root

Answer 6

i know, just curios.

Answer 7

-1

Answer 8

...

Answer 9

i thought that was the square... exponent 2.

Answer 10

-1*-1=1 not i.

Answer 11

\[\Large i = \sqrt{-1}\]
\[\Large i^2 = (\sqrt{-1})^2\]
\[\Large i^2 = -1\]

Answer 12

\[\iota ^{2} = -1\]

Answer 13

am i just the stupid idiot here or u genuinely dont know or connection problemos?

Answer 14

square root. sry like radical.

Answer 15

sorry @Plasmataco I'm not following

Answer 16

i know but like -1 under a radical of 4

Answer 17

were you able to find r and theta, @lxoser ?

Answer 18

...

Answer 19

oh you mean

\[\Large \sqrt{i} = \sqrt[4]{-1}\] @Plasmataco ??

Answer 20

\[\sqrt[4]{-1}\]

Answer 21

yeah

Answer 22

im gonna need help finding r and theta @jim_thompson5910

Answer 23

sry for interrupting...

Answer 24

its okay

Answer 25

\[\Large r = \sqrt{a^2 + b^2}\]
\[\Large r = \sqrt{3^2 + (3\sqrt{3})^2}\]
...
...
...
\[\Large r = ??\]

Answer 26

r = 30?

Answer 27

dont think so.

Answer 28

simplify.

Answer 29

it should be somthing a lot lower

Answer 30

Hint:
\[\Large (3\sqrt{3})^2=(3\sqrt{3})*(3\sqrt{3})\]
\[\Large (3\sqrt{3})^2=(3*3)*(\sqrt{3}*\sqrt{3})\]
\[\Large (3\sqrt{3})^2=(3*3)*\sqrt{3*3}\]
\[\Large (3\sqrt{3})^2=9\sqrt{9}\]
\[\Large (3\sqrt{3})^2=9*3\]
\[\Large (3\sqrt{3})^2=27\]

Answer 31

r = 6

Answer 32

yes

Answer 33

yup.

Answer 34

horay! now my question.

Answer 35

for theta, i got -60?

Answer 36

incorrect

Answer 37

is the result suppose to be in radians or degrees?

Answer 38

it depends on what the instructions say

Answer 39

does it say which mode they want?

Answer 40

no, but these are the answer choices

Answer 41

ok so they want radian form

Answer 42

so would theta be - pi/3

Answer 43

yes, now because our point is in Q2 (see graph) we add pi radians to our angle to land in the right quadrant

Answer 44

- pi/3 is in Q4