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

@amistre64
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 2

@phi
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 3

Did they teach you the method to do this ?

Answer 4

no actually i am taching miself through your help openstudy
Could you help me out

Answer 5

because I would write both in the form \( A e^{i \theta} \)

Answer 6

can you show me how to do it

Answer 7

Euler's formula http://en.wikipedia.org/wiki/Euler%27s_formula
\[ e^{ ix} = \cos x + i \sin(x) \]
so
\[ z_1= \sqrt{3} \left(\cos\left( \frac{7\pi}{4}\right) + i \sin\left( \frac{7\pi}{4}\right)\right) = \sqrt{3} e^{i \frac{7\pi}{4}}\]
and
\[ z_2= \sqrt{6} e^{i \frac{9\pi}{4}}\]
so
\[ \frac{z_1}{z_2}= \frac{\sqrt{3} e^{i \frac{7\pi}{4}}}{\sqrt{6} e^{i \frac{9\pi}{4}}} \]

Answer 8

i have a question @phi What does "i" stand for?

Answer 9

i = \( \sqrt{-1} \)

as you know, there is no "normal" i.e. real number that when you square it, gives -1.
1*1= 1
and -1 * -1 = 1
so people said, let's just say there is such a number (and because sqrt(-1) is awkward to write out all the time) they said, we will call it i
i*i= -1 by definition

You might be able to follow Khan's great video on this
https://www.khanacademy.org/math/calculus/sequences_series_approx_calc/maclaurin_taylor/v/euler-s-formula-and-euler-s-identity

Answer 10

to finish, use the "exponent rule"
\[ \frac{a^b}{a^c} = a^{b-c} \]
\[ \frac{z_1}{z_2}= \frac{\sqrt{3} e^{i \frac{7\pi}{4}}}{\sqrt{6} e^{i \frac{9\pi}{4}}} =
\frac{\sqrt{3}}{\sqrt{6}} e^{- \frac{\pi}{2}}= \frac{\sqrt{2}}{2}\left( \cos\left(- \frac{\pi}{2}\right) +i \sin\left(- \frac{\pi}{2}\right)\right)
\]
we can simplify that using cos(-x)= cos(x) and sin(-x)= - sin(x)
\[ \frac{\sqrt{2}}{2}\left( \cos\left(\frac{\pi}{2}\right) -i \sin\left(\frac{\pi}{2}\right)\right)
\]
of course we know cos(pi/2) is 0 and sin(pi/2) = 1
so we can simplify further