Question

Find all solutions of sin(z)=0?

Answer 1

all of them? and is z the complex variable or just some random variable that is real?

Answer 2

z= complex

Answer 3

every odd iteration of pi makes sin(x) = 0 ..... but Z eh

Answer 4

thought I erased that odd part lol

Answer 5

z=2*k*pi

Answer 6

z=2kpi? Where K is a real no.? :D

Answer 7

\[k \in Z\]

Answer 8

I dont know of any sin(z) =0 that exists in the complex part

Answer 9

Idk how u may call them...Sorry...

Answer 10

if we start out as the [0,inf) real line as our initial ray, then any tine a + 0i occurs we get sin=0

Answer 11

so at 1+0i and -1+0i times whatever scalar you wanna used for the unit circle

Answer 12

But a+0i isn't a complex number, So you're saying every integer is a complex number, aren't you?

Answer 13

every integer is a complex number in the same sense that every integer is a rational number; because they are inherently a part of the set..

Answer 14

the complex plane includes the real number line; so yes... all real numbers are a subset of the complex plane

Answer 15

OK, if so, then the answer is just one, which is 0.

Answer 16

\[R \subset complex no.s\]

Answer 17

its only '1' if we make a unit circle in the complex plane and choose the +Reals as our initial ray; then rotate around and terminate everytime we hit the real line again...

Answer 18

when Z = any complex number of the form: a+0i; we get sin(z) = 0

Answer 19

But only integer which is of the form npi is 0

Answer 20

So the answer should be just one

Answer 21

But now what confuses me is, x-1=0 has a complex solution?

Answer 22

Z is not retricted to "off of the real number line" so just sating '1' is useless

Answer 23

Not 1....4 example...k=-1---> Z=2(-1)*pi=-2pi=angle 0
k=0--->Z=angle 0
k=1--->=2*1*pi=angle 0

Answer 24

the coordinate system is like the reals; but you define Z as: z = (x + yi)

Answer 25

as opposed to: point = (x,y)

Answer 26

Oh, so a+bi is a complex number,
so is a + 0i
so is R+0i

Thus the solution to this would be npi+0i, wouldn't it?

Answer 27

exactly :)

Answer 28

but to say npi means to say only those z values that are a real multiple of pi is wrong

Answer 29

every point along the real number line is valid

Answer 30

pi is a number of the real number line; its a measure of the diatmer of a circle to its circumference

Answer 31

Is the polar form correct, bcuz you are saying sin(1) = 0?

Answer 32

sin(1 +0i) = 0 at least :)

Answer 33

sin(1+0i) = 0?

Answer 34

sin(z) = 0 means at least:

sin(a + bi) = 0

Answer 35

sin(1)=0....? how come?

Answer 36

sin(a + 0i) = 0 for any given a

Answer 37

But polar form just gives you a way to express

Answer 38

Not a complex plane

Answer 39

polar form can be used to interpret the complex plane into a real plane;

Answer 40

Hmm., check this out http://www.wolframalpha.com/input/?i=sin%281%2B0i%29

Answer 41

it wold resemble the polar form yes...

Answer 42

Don't take me wrong PLEASE....but....why do u explain a very easy problem in this way?....O.o....ur making it harder then it really is....O.o

Answer 43

than*

Answer 44

Don't you think, we are leaving out someone in this discussion? The asker!

Answer 45

i havent blocked them lol

Answer 46

wolfram is good but still has issues with interpreting data

Answer 47

haha lol :D.....maybe he has got it...:)

Answer 48

hey..i'm still here,get confuse
@_@

Answer 49

Oops....@Amistre do u agree with me that the answer is Z=2kpi?

Answer 50

im open to me being wrong :) but I would have to see how sin(z) is sposed to be interpreted :)

Answer 51

K belongs to Z

Answer 52

i'm reading this post, try to digest, what do u guys think:
http://answers.yahoo.com/question/index?qid=20080529224140AAGdnOg

Answer 53

angela, I would have to research it to tell :)

Answer 54

Ok.....I have always thought math is same everywhere...but...o.O

Answer 55

maybe sin(z) = 0 where e^(ix) = e^(-ix) :)

Answer 56

maybe....ur the smart guy here anyway :P

Answer 57

u have Z={..........-3,-2,-1,0,1,2,3.........} right?

Answer 58

no; z is beig used in this context to indicate the complex variable and not the set of integers

Answer 59

its like using x for a real variable;

they use z for the complex variable

Answer 60

Really??????? well in my answer i meant for this kind of Z.....and wht do u call ''my Z''?

Answer 61

but all signs point to the answer being npi where n is any integer

Answer 62

Integers!

Answer 63

your Z would indicate "any integer" ;

Answer 64

ahaa...that's wht i meant....:S:S

Answer 65

Thanks :):):) ^_^

Answer 66

Using 0 = (e^(iz) - e^(-iz)) / 2i

I get stucked when
2iz= log 1

what's next...

Answer 67

@angela210793: You never seemed to be not knowing that though! lol

Answer 68

http://jwbales.us/precal/part6/part6.3.html

this is more my thinking

Answer 69

sin(z) = bi/|z|

Answer 70

when bi=0; sin(z) = 0

Answer 71

Hey...X( X( I know what that is X( is just that i thought integers in english was Z as in my place is...

Answer 72

I saw one textbook have it as J and I was like... whats that? lol

Answer 73

Why can't they keep it to just one universal symbol~!

Answer 74

37 mins ago I've written:''Idk how u may call them...Sorry...''
So it's not my fault..:(:(:(

Answer 75

ho ho ho

Answer 76

i blame the chinese... cant trust anyone theat squints all the time..shifty beady little eyes lol

Answer 77

hahahaha :D :D Lool :D :)

Answer 78

\[z=k\pi, k\in Z\]

Answer 79

what is with the latex???

Answer 80

O.o

Answer 81

its new since they upgraded today; nothing works lol,

Answer 82

@satellite73, are u confident with that answer?
which method u're using?