Question

Find the quotient z1/z2 of the complex numbers. Leave answer in polar form.
z1= sqrt3 (cos 7pie/4 + i sin 7pie/4)
z2 = sqrt6 (cos 9pie/4 + i sin 9pie/4)

Answer 1

divide the numbers, subtract the angles

Answer 2

i am really lost with this one

Answer 3

\[\frac{\sqrt{3}}{\sqrt{6}}\left(\cos(\frac{(7-9)\pi}{4})+i\sin(\frac{(7-9)\pi}{4})\right)\]

Answer 4

is that the answer or is there another step

Answer 5

it is supposed to be easy, that is why it is used
to divide, take \(\frac{r_1}{r_2}\) for the absolute value, and subtract the angles for the new angle

Answer 6

yes, there is another step, you should compute
i only wrote what you need to compute

Answer 7

do you know what you need your calculator in to do this

Answer 8

\(\frac{7\pi}{4}-\frac{9\pi}{4}=-\frac{2\pi}{4}=-\frac{\pi}{2}\) is your angle

Answer 9

no calculator needed

Answer 10

and \[\frac{\sqrt{3}}{\sqrt{6}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\]

Answer 11

so is it sqrt2/2 (cos pie/2 + i sin pie/2)

Answer 12

yup

Answer 13

oh no!!

Answer 14

angle is minus

Answer 15

the sqrts confuse me

Answer 16

\[\frac{\sqrt{2}}{2}\left(\cos(-\frac{\pi}{2})+\sin(-\frac{\pi}{2})\right)\]

Answer 17

okay thank you

Answer 18

i forgot the \(i\) it is
\[\frac{\sqrt{2}}{2}\left(\cos(-\frac{\pi}{2})+i\sin(-\frac{\pi}{2})\right)\]

Answer 19

please note that this is easy. one division, one subtraction

Answer 20

yw

Answer 21

i have three more can you help me I got them wrong the first time and I want to see what I did wrong

Answer 22

post in a new post and i will look at them

Answer 23

hard to keep scrolling down here

Answer 24

okay