Question

Need help dividing complex number into polar form

Answer 1

I got up to 5cis(2pi/3-pi/4) but idk how 2pi-pi = 5pi?

Answer 2

v^-1 and then you can add the angles

v^-1 = (1/2M)cis(-pi/4)

Answer 3

w/v = w*(v^-1)

Answer 4

Sorry, I don't understand?

Answer 5

\[v^{-1}=\frac{ 1 }{ 2M }cis \left( -\frac{ \pi }{ 4 } \right)\]
\[\frac{ w }{ v }=w\cdot v^{-1}=10M cis \left( \frac{ 2\pi }{ 3 } \right)\cdot \frac{ 1 }{ 2M }cis \left( -\frac{ \pi }{ 4 } \right)\]

Answer 6

Yeah, and I got 5cis (2pi/3 - pi/4)

Answer 7

and the answer is 5cis(5pi/4) but I don't understand how to get the 5pi?

Answer 8

cis? you mean cos right

Answer 9

\[5cis \left( \frac{ 2\pi }{ 3 }-\frac{ \pi }{ 4 } \right)=5 cis \left( \frac{ 8\pi }{ 12 }-\frac{ 3\pi }{ 12 } \right)\]

Answer 10

@UsukiDoll The question says cis

Answer 11

what?

Answer 12

oh wow gotta be a typo

Answer 13

cos + i sin

Answer 14

cis = cos + i sin

Answer 15

ha ha very funny

Answer 16

cuz I sign x)

Answer 17

@pgpilot326 why is 2pi and pi equal to 8pi and 3pi?

Answer 18

\[w=10M cis \left( \frac{ 2 \pi }{ 3 } \right)=10m \left( \cos \left( \frac{ 2 \pi }{ 3 } \right)+i \sin \left( \frac{ 2 \pi }{ 3 } \right) \right)\]

Answer 19

\[v^{-1}=\frac{ 1 }{ 2m }cis\left( -\frac{ \pi }{ 4 } \right)= \frac{ 1 }{ 2m }\left( \cos\left( -\frac{ \pi }{ 4 } \right)+i \sin\left( -\frac{ \pi }{ 4 } \right) \right)\]

Answer 20

when you multiply complex numbers in polar form, you multiply the magnitudes of the vectors and add the angles. in order to add fractions you need a common denominator.

do you follow?

Answer 21

Yeah I think

Answer 22

https://www.khanacademy.org/math/trigonometry/imaginary_complex_precalc/complex_analysis/e/multiplying_and_dividing_complex_number_polar_forms

Answer 23

Thanks