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)
divide the numbers, subtract the angles
i am really lost with this one
\[\frac{\sqrt{3}}{\sqrt{6}}\left(\cos(\frac{(7-9)\pi}{4})+i\sin(\frac{(7-9)\pi}{4})\right)\]
is that the answer or is there another step
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
yes, there is another step, you should compute i only wrote what you need to compute
do you know what you need your calculator in to do this
\(\frac{7\pi}{4}-\frac{9\pi}{4}=-\frac{2\pi}{4}=-\frac{\pi}{2}\) is your angle
no calculator needed
and \[\frac{\sqrt{3}}{\sqrt{6}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\]
so is it sqrt2/2 (cos pie/2 + i sin pie/2)
yup
oh no!!
angle is minus
the sqrts confuse me
\[\frac{\sqrt{2}}{2}\left(\cos(-\frac{\pi}{2})+\sin(-\frac{\pi}{2})\right)\]
okay thank you
i forgot the \(i\) it is \[\frac{\sqrt{2}}{2}\left(\cos(-\frac{\pi}{2})+i\sin(-\frac{\pi}{2})\right)\]
please note that this is easy. one division, one subtraction
yw
i have three more can you help me I got them wrong the first time and I want to see what I did wrong
post in a new post and i will look at them
hard to keep scrolling down here
okay
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