Question

how do you solve z1=4i
z2=-6+6i

Answer 1

I'd try my hand at this if I knew the context of this question haha

Answer 2

there is nothing to solve for

Answer 3

it says find quotient z1/z2 of the complex numbers leave the answer in polar form

Answer 4

now there is something to do

Answer 5

@johnbc that is just it

Answer 6

and if you are working in polar form the division is easy, all the work is writing \[4i\] ad \[-6+6i\] in polar form

Answer 7

do you know how to do that?

Answer 8

if the answer is "no" that is fine, we can do them, they are not hard in this case
just asking is all

Answer 9

I started it out and this is what I got r1=4^2=4
r2=\[\sqrt{-6^2+6i^2}\]=6\[\sqrt{2}\]

Answer 10

k good work

Answer 11

so \(r=6\sqrt2\) how about \(\theta\) ?

Answer 12

i meant
\[r_1=4,r_2=6\sqrt2\] like you said
btw you do not need a formula to know that \(|4i|=4\) it is obvious
but the formula does work

Answer 13

tan\[\tan \theta \](6/-6)

Answer 14

that is one way to do it
another it to look at the picture

Answer 15

that angle should be more or less obvious
you working in radians (i hope) or degrees?

Answer 16

degrees

Answer 17

k then does the picture give the answer?

Answer 18

on the test I need to know how to solve it with the formula can you help me with that this is in my review packet

Answer 19

ok but really you are going to need the picture in any case
you wrote
\[\tan(\theta)=\frac{6}{-6}\] which is
\[\tan(\theta)=-1\]

Answer 20

yes

Answer 21

so how to find \(\theta\) ?
you could try \(\tan^{-1}(-1)\) but that will not actually work
that will give you \(-45\) and that is in the wrong quadrant
so you pretty much have to know from the picture that the angle is \(135\)

Answer 22

yes I was given the formula z1/z2=r1/r2[cos\[\theta1\]-\[\theta2\]+I isin(\[\theta1-\theta2\])

Answer 23

yeah i got that
first we need to know that
\[4i=4(\cos(90)+i\sin(90))\]and
\[-6+6i=6\sqrt2(\cos(135)+i\sin(135))\]

Answer 24

like is said, all the work is writing them in polar form
the division is easy

Answer 25

okay with you so far

Answer 26

are you good with this one \[-6+6i=6\sqrt2(\cos(135)+i\sin(135))\]

Answer 27

I have the answer key and it is not matching up

Answer 28

not sure what you mean
we are not done yet

Answer 29

for the tw the teach said it should be \[\sqrt{2}/3\](cos7pi/4 +I sin 7pi/4)

Answer 30

yes because we are not done
also you lied to me!! you said you were using degrees, but you are using radians

Answer 31

sorry thought we were

Answer 32

k lets review the steps
you are not finished which is why we do not have the teachers answer
you have to write both
\[4i\] and \[-6+6i\] in polar form
i .e. write
\[4i=r(\cos(\theta)+i\sin(\theta))\] and
\[-6+6i=s(\cos(\alpha)+i\sin(\alpha))\] we have not done either of those yet
so far all we know is
\[r=4,s=6\sqrt2\]

Answer 33

yes

Answer 34

so we need \(\theta\) and \(\alpha\)

the picture should help you see that \(\alpha =\frac{3\pi}{4}\) which previously i wrote as \(135\)

Answer 35

okay

Answer 36

you also need \(\theta\) which should be clear if you plot \(4i\)

Answer 37

did you get \(\theta =\frac{\pi}{2}\) ?

Answer 38

no how did you get that my calc gives me undefined

Answer 39

again you have to know where you are

Answer 40

okay I get it

Answer 41

that angle is clearly \(90\) or \(\frac{\pi}{2}\)

Answer 42

you are trying to use a formula
that is fine but you should not get married to it
look to see where you are


now we can finish easy