Question

any precal tutors out there!!!??

Answer 1

Yes, what's up?

Answer 2

@amistre64

Answer 3

id say its gonna at least be a 4/3 but thats just a guess at the moment

Answer 4

ok

Answer 5

conjugates should get you to a happier place is another thought i have .... no matter how wrong it may be :)

Answer 6

hmmmm

Answer 7

isnt there some e^(-??) form they can equate into to make this simpler?

Answer 8

(a1 + ib1) + (a2 + ib2) = (a1 + a2) + i(b1 + b2).

Answer 9

he complex conjugate of a complex number z = a + ib is denoted by and defined as:

.

Answer 10

but we are dividing, not adding

Answer 11

conjugates just swap out the operation + for - and visaversa

Answer 12

(a+b)(a-b) = a^1-b^2

Answer 13

oh cool, it looks like i was on the right track at least with the conjugate idea :)

Answer 14

ya so i dont get what to plug in to that

Answer 15

if you could type in the problem into here, it would be eaiser for me to play with. I get lost swapping back and forth between windows

Answer 16

sure!

Find the quotient of z1 by z2. Express your answer in trigonometric form.
z1= 4(cos(pi/3) + i sin(pi/3) )

z2= 3(cos(2pi/5) + i sin(2pi/5) )

Answer 17

\[e^{ix}=cosx+isinx\]So now you can convert this into a simple algebra exponent rules problem.\[e^{i \pi/3}/e^{i2\pi/5}=e^{i(\pi/3-2\pi/5)}=e^{i(-\pi/15)}\]

Answer 18

4(cos(pi/3) + i sin(pi/3) )(cos(2pi/5) + i sin(2pi/5) )
3(cos(2pi/5) + i sin(2pi/5) )(cos(2pi/5) + i sin(2pi/5) )

4(cos(pi/3)cos(2pi/5) + i sin(pi/3)cos(2pi/5)
+ cos(pi/3)i sin(2pi/5) + i sin(pi/3)i sin(2pi/5))

3(cos^2(2pi/5) + sin^2(2pi/5) )

4(cos(pi/3)cos(2pi/5) + i sin(pi/3)cos(2pi/5)
+ cos(pi/3)i sin(2pi/5) - sin(pi/3)sin(2pi/5))
3(2pi/5 )


Kainuis got the way easier method :)

Answer 19

@Kainui can you help me plug that in

Answer 20

only one option has the 4/3 witha -pi/15 in it

Answer 21

oh i see now!! ok thanks guys i have a couple more..

Answer 22

post them to the left, that way thisone doesnt get bogged down :)

Answer 23

This is exactly the same problem as the last one. You will use the exact same equation, e^(ix)=cosx+isinx

Answer 24

degree*pi/180 converts these to radians

Answer 25

i dont see a 5/2 option tho ....

Answer 26

Honestly you only really need to know how to multiply the coefficients here. The difference is this says find the product not the quotient, amistre64.

Answer 27

doh!!

Answer 28

lol

Answer 29

so i plug them in to the formula?

Answer 30

to use the formula, if you must, youd have to convert from degrees to radians, and then change back again in the end

Answer 31

any way around that?

Answer 32

its simple enough to do, itd be like trying to go thru alaska to get from florida to georgia

Answer 33

\[\Large 5e^{i\frac{25pi}{180}}*2e^{i\frac{80pi}{180}}=10e^{i\frac{105pi}{180}}\]

Answer 34

this one?

Answer 35

e^(ix)=cosx+isinx

Answer 36

i spose just adding the degrees would amount to the same thing ....

Answer 37

so multiply?? those out

Answer 38

i learn by doing :)

Answer 39

Nope, see all you're doing is looking at cosx+isinx and seeing that they have the same value for x. Then you take that x and plug it into e^(ix). So for example if you have:

cos2+isin2 your x=2 right? So then you can rewrite that all as e^(i2).

Answer 40

what if their not the same?

Answer 41

then they would be a "z" prolly