Question

Find the product of z1 and z2 where z1 = 8(cos 40° + i sin 40°), and z2 = 4(cos 135° + i sin 135°).

OPTIONS:
32(cos 40° + i sin 40°)

12(cos 175° + i sin 175°)

32(cos 5400° + i sin 5400°)

32(cos 175° + i sin 175°)

PLEASE HELP PLEASE

Answer 1

Hint:

p*cis(x) times q*cis(y) = (p*q)*cis(x+y)

where

cis(x) = cos(x) + i*sin(x)

Answer 2

I dont get it :(

Answer 3

@jim_thompson5910

Answer 4

z1*z2

[8(cos 40° + i sin 40°)] * [4(cos 135° + i sin 135°)]

[8*cis(40)] * [4*cis(135)]

does that help?

Answer 5

so its A right @jim_thompson5910

Answer 6

no

Answer 7

z1*z2

[8(cos 40° + i sin 40°)] * [4(cos 135° + i sin 135°)]

[8*cis(40)] * [4*cis(135)]

(8*4)*cis(40+135) ... use the formula I gave you

See the answer now?

Answer 8

Its D then @jim_thompson5910

Answer 9

correct

Answer 10

One more ?

Answer 11

Find the cube roots of 27(cos 279° + i sin 279°).
Could you help with this
@jim_thompson5910

Answer 12

use the formula found on this page

http://orion.math.iastate.edu/trig/sp/xm08/applets/complexcuberoot.html

Answer 13

AHH @jim_thompson5910 CONFUSION

Answer 14

all you're doing is plugging in k = 0, k = 1, k = 2 to find the 3 cube roots

Answer 15

r = 27

theta = 279

Answer 16

hopefully this isn't brand new to you

Answer 17

let me try

Answer 18

ok

Answer 19

^zk= ^3sqrt 27(cos (279 +2kpi/3) + isin (279+2kpi/3))
@jim_thompson5910

Answer 20

perfect

Answer 21

now to find the 3 cube roots, you substitute k = 0, k = 1, and k = 3

you do each value of k one at a time

Answer 22

That I dont know how to do

Answer 23

so for example, when k = 0, the first cube root is

z0 = 3*(cos(279+2*0*pi/3) + i*sin(279+2*0*pi/3))

z0 = 3*(cos(279/3) + i*sin(279/3))

z0 = 3*(cos(93) + i*sin(93))

So that is your first cube root

Answer 24

plug in k = 1 to find the next cube root

Answer 25

okay

Answer 26

tell me what you get

Answer 27

wouldn't it be 3(-sin(279+pi/6) + icos(279+pi/6)

Answer 28

? @jim_thompson5910

Answer 29

close

z1 = 3*(cos(279+2*1*pi/3) + i*sin(279+2*1*pi/3))


z1 = 3*(cos( 279+2pi/3) + i*sin(279+2pi/3))

Answer 30

and I'd just leave it like that

Answer 31

And the last
let me try

Answer 32

eek

Answer 33

is it just 3(cos (279) + isin (279)) @jim_thompson5910

Answer 34

not quite

Answer 35

unfortunately it's not that clean cut

Answer 36

:(

Answer 37

you plugged in k = 2 right?

Answer 38

3 right ?

Answer 39

Im confused about this one help:( sorry

Answer 40

no k = 2

Answer 41

I thought I only am supposed to plug in k = 0, k = 1, and k = 3

Answer 42

no, k = 0, k = 1, k = 2

Answer 43

@jim_thompson5910

Answer 44

like the page describes

Answer 45

z2 = 3*(cos(279+2*2*pi/3) + i*sin(279+2*2*pi/3))

z2 = 3*(cos(279+4pi/3) + i*sin(279+4pi/3))

and i'd leave it like that

Answer 46

THATS IT I had a different answer THANK YOU SO MUCH
is there anything else ?

Answer 47

nope, we found all 3 roots using the correct formula

Answer 48

so we're done

Answer 49

There is your medal as promised :)

Answer 50

can i ask what happened to the 27 in the problem? it was in the formula then it seemed like it disappeared .. @jim_thompson5910