Question

Please Help!!

Express the complex number in trigonometric form.

3i

Answer 1

if u have complex number
z=x+iy
then u can write it as
z= r(cos theta + i sin theta )

such that r=|z|

Answer 2

3i => 0 + 3i
x y
thus

Answer 3

\(\large \bf z= r[cos(\theta)+isin(\theta)]\)

Answer 4

im not understanding

Answer 5

well.. you'd need to find keep in mind that the complex a + bi form correlates to the rectangular (x,y) coordinates

from there, you get the magnitude or "r" or hypotenuse if you like
and the angle

Answer 6

what would be the "r" or magnitude of that vector?
what's the angle?

Answer 7

thus \(\large \bf {\color{blue}{ r}}[cos({\color{purple}{ \theta}})+isin({\color{purple}{ \theta}})]\)

Answer 8

is r 3?

Answer 9

x = 0 y = 3 \(\bf {\color{blue}{ r}}=\sqrt{a^2+b^2}\to \sqrt{x^2+y^2}\to \sqrt{0^2+3^2}\to 3\)

Answer 10

ok , r = length of z u can find it through
\(r=\sqrt (x^2 +y^2)\)
and as
if u have complex number
z=x+iy then u can write it as
z= r(cos theta + i sin theta ) such that r=|z|


then x= r cos theta
y= r sin theta


then find theta , r :D

Answer 11

ok, so r is 3

Answer 12

and the angle, should be quite obvious

Answer 13

90

Answer 14

can i ask for help on another one?

Answer 15

yeap, thus \(\large \bf {\color{blue}{ 3}}[cos({\color{purple}{ 90^o}})+isin({\color{purple}{ 90^o}})]\)

Answer 16

Express the complex number in trigonometric form.

-3 + 3 (sqrt 3)i

Answer 17

same way really

Answer 18

so r is 3sqrt3?

Answer 19

\(\large \begin{array}{cccllll}
-3&+3\sqrt{3}\ i\\
\uparrow &\uparrow \\
x&y
\end{array}\)

Answer 20

\(\bf {\color{blue}{ r}}=\sqrt{a^2+b^2}\to \sqrt{x^2+y^2}\)

Answer 21

so 6

Answer 22

how would i find the angle now?

Answer 23

yeap

so you'd find the "r" or magnitude

and to find the angle... keep in mind that \(\bf tan(\theta)=\cfrac{y}{x}\to \cfrac{3\sqrt{3}\ i}{-3}\quad thus\quad {\color{purple}{ \theta}}=tan^{-1}\left(\cfrac{3\sqrt{3}\ i}{-3}\right)\)

Answer 24

oh ok

Answer 25

well... I should have removed the "i' from there =) \(\bf tan(\theta)=\cfrac{y}{x}\to \cfrac{3\sqrt{3}}{-3}\quad thus\quad {\color{purple}{ \theta}}=tan^{-1}\left(\cfrac{3\sqrt{3}}{-3}\right)\)

Answer 26

so if i change that to cos and i sin would that be 6(cos (2pi/3) + i sin (2pi/3))

Answer 27

wait no, it should be (5pi/6)

Answer 28

yeap

well.. keep in mind that tangent is negative in II and IV quadrants

Answer 29

can you help me with one more? im half way through with it, but i don't know how to finish it

Answer 30

Find the cube roots of 125(cos 288° + i sin 288°).
cubed root(125*(cos(288)+ i sin (288))
cubed root(125)*cubed root(cos(288)+i sin(288))
5*(cos(288)+ i sin(288))^1/3

Answer 31

actaully should be the 1st one you listed \(\bf \cfrac{2\pi}{3} \quad and\quad \cfrac{5\pi}{3}\)

Answer 32

oh ok

Answer 33

is simpler posting anew... if I dunno someone else may, and thus more exposure and also we can reviise each other :)

Answer 34

ok :)