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Mathematics 30 Online
OpenStudy (anonymous):

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)

OpenStudy (anonymous):

divide the numbers, subtract the angles

OpenStudy (anonymous):

i am really lost with this one

OpenStudy (anonymous):

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

OpenStudy (anonymous):

is that the answer or is there another step

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

no calculator needed

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

oh no!!

OpenStudy (anonymous):

angle is minus

OpenStudy (anonymous):

the sqrts confuse me

OpenStudy (anonymous):

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

OpenStudy (anonymous):

okay thank you

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yw

OpenStudy (anonymous):

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

OpenStudy (anonymous):

post in a new post and i will look at them

OpenStudy (anonymous):

hard to keep scrolling down here

OpenStudy (anonymous):

okay

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