Question

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

32(cos 40° + i sin 40°)
12(cos 175° + i sin 175°)
32(cos 5400° + i sin 5400°)
32(cos 175° + i sin 175°)

Answer 1

@amistre64 @Nnesha

Answer 2

theres a general rule, but it seems that you dont know the rule

Answer 3

multiply it out ....

Answer 4

like distribute?

Answer 5

yeah, its basic algebra, with a trig twist :)

Answer 6

n (cos(a) + i sin(a)) * m (cos(b) + i sin(b))

nm * (cos(a) + i sin(a)) (cos(b) + i sin(b))

the first part of the rule is that "radiuses" multiply so with that we can focus on the trigs

Answer 7

simplify for the sake of typing

(cos(a) + i sin(a)) (cos(b) + i sin(b))
r s p q

(r + is)(p + iq)

Answer 8

32 X (r+ is)(p+iq)

Answer 9

correct, now expand the right side

Answer 10

(cos 40° + i sin 40°)(cos 135° + i sin 135°)

Answer 11

(r+ is)(p+iq)

r(p+iq)+ is(p+iq)

keep it going

Answer 12

cos40 (cos 135 + i sin 135)

Answer 13

we need to keep this as general as possible so that you can see how the stuff works. and not be blind sided by your specific problem values.

r(p+iq) + is(p+iq)

rp + i rq + i sp + ii sq
^^
ii = -1

rp + i rq + i sp - sq

collect like terms, real to real, and i to i

(rp - sq) + i (rq + sp)

now we can replace them with the general trigs we subbed for, recall:

r s p q
(cos(a) + i sin(a)) (cos(b) + i sin(b))


real part is: rp - sq
imaginary part is: rq + sp

real part is: cos(a)cos(b) - sin(a)sin(b)
imaginary part is: cos(a)sin(b) + sin(a)cos(b)

these trigs are identities for something, do you recall them?

Answer 14

we are constructing the general rule for these types of problems, so that we dont have to work it out the long way each and every time

Answer 15

the identities are the sum and difference formulas

Answer 16

well, the sum formulas yes

real part is: cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
imaginary part is: sin(a+b) = cos(a)sin(b) + sin(a)cos(b)

therefore the rule is

the radiuses multiply, the arguments add

Answer 17

ok

Answer 18

\[n(\cos(a)+i\sin(a))~*~m(\cos(b)+i\sin(b))\implies nm(\cos(a+b)+i\sin(a+b))\]

Answer 19

so the 4th option? 32(cos 175° + i sin 175°)

Answer 20

in a more succinct notation:
\[(n,a)(m,b)\implies (nm,a+b)\]
\[(8,40)(5,135)\implies (32,175)\]yes

Answer 21

thanks

Answer 22

also do you know how to do the cube roots? I forgot how to do the degrees

Answer 23

yeah, i can muddle my way thru it

Answer 24

do you want a specific example?

Answer 25

\[(n,a)(n,a)\implies (n^2,2a)\]
\[(n,a)(n,a)(n,a)\implies (n^2,2a)(n,a)\implies(n^3,3a)\]

as a rule

\[(n,a)^k\implies (n^k,ka)\]

Answer 26

Find the cube roots of 27(cos 279° + i sin 279°).

Answer 27

cube root is equivalent to ^(1/3)

Answer 28

yes, but I need to know how to do the degree portion. I know the 3rd root of 27 is 3

Answer 29

ive already posted the rule ....

Answer 30

oh

Answer 31

so 3rdrt of 279 and 279/3?

Answer 32

that is the "primitive" degree for lack of a better term.

we add (360/3) to it to find all 3 roots and adjust as needed

Answer 33

so 279+120 and -120?

Answer 34

correct

Answer 35

actually it would be -120 and -240

Answer 36

279/3 + 120(0)
279/3 + 120(1)
279/3 + 120(2)

and adjust as needed to keep them between 0 and 360

Answer 37

actually it would be -120 and -240

Answer 38

7th roots would be?

279/7 + 360(0)/7
279/7 + 360(1)/7
279/7 + 360(2)/7
279/7 + 360(3)/7
279/7 + 360(4)/7
279/7 + 360(5)/7
279/7 + 360(6)/7

3^(1/7) of course

Answer 39

a more convenient way to view this is to divide the circle into its parts, and then add the argument to each part to rotate it appropriately .. does that make sense?

Answer 40

yes