Find all solutions of sin(z)=0?

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Find all solutions of sin(z)=0?

Mathematics
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all of them? and is z the complex variable or just some random variable that is real?
z= complex
every odd iteration of pi makes sin(x) = 0 ..... but Z eh

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Other answers:

thought I erased that odd part lol
z=2*k*pi
z=2kpi? Where K is a real no.? :D
\[k \in Z\]
I dont know of any sin(z) =0 that exists in the complex part
Idk how u may call them...Sorry...
if we start out as the [0,inf) real line as our initial ray, then any tine a + 0i occurs we get sin=0
so at 1+0i and -1+0i times whatever scalar you wanna used for the unit circle
But a+0i isn't a complex number, So you're saying every integer is a complex number, aren't you?
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..
the complex plane includes the real number line; so yes... all real numbers are a subset of the complex plane
OK, if so, then the answer is just one, which is 0.
\[R \subset complex no.s\]
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...
when Z = any complex number of the form: a+0i; we get sin(z) = 0
But only integer which is of the form npi is 0
So the answer should be just one
But now what confuses me is, x-1=0 has a complex solution?
1 Attachment
Z is not retricted to "off of the real number line" so just sating '1' is useless
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
the coordinate system is like the reals; but you define Z as: z = (x + yi)
as opposed to: point = (x,y)
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?
exactly :)
but to say npi means to say only those z values that are a real multiple of pi is wrong
every point along the real number line is valid
pi is a number of the real number line; its a measure of the diatmer of a circle to its circumference
Is the polar form correct, bcuz you are saying sin(1) = 0?
sin(1 +0i) = 0 at least :)
sin(1+0i) = 0?
sin(z) = 0 means at least: sin(a + bi) = 0
sin(1)=0....? how come?
sin(a + 0i) = 0 for any given a
But polar form just gives you a way to express
Not a complex plane
polar form can be used to interpret the complex plane into a real plane;
Hmm., check this out http://www.wolframalpha.com/input/?i=sin%281%2B0i%29
it wold resemble the polar form yes...
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
than*
Don't you think, we are leaving out someone in this discussion? The asker!
i havent blocked them lol
wolfram is good but still has issues with interpreting data
haha lol :D.....maybe he has got it...:)
hey..i'm still here,get confuse @_@
Oops....@Amistre do u agree with me that the answer is Z=2kpi?
im open to me being wrong :) but I would have to see how sin(z) is sposed to be interpreted :)
K belongs to Z
i'm reading this post, try to digest, what do u guys think: http://answers.yahoo.com/question/index?qid=20080529224140AAGdnOg
angela, I would have to research it to tell :)
Ok.....I have always thought math is same everywhere...but...o.O
maybe sin(z) = 0 where e^(ix) = e^(-ix) :)
maybe....ur the smart guy here anyway :P
u have Z={..........-3,-2,-1,0,1,2,3.........} right?
no; z is beig used in this context to indicate the complex variable and not the set of integers
its like using x for a real variable; they use z for the complex variable
Really??????? well in my answer i meant for this kind of Z.....and wht do u call ''my Z''?
but all signs point to the answer being npi where n is any integer
Integers!
your Z would indicate "any integer" ;
ahaa...that's wht i meant....:S:S
Thanks :):):) ^_^
Using 0 = (e^(iz) - e^(-iz)) / 2i I get stucked when 2iz= log 1 what's next...
@angela210793: You never seemed to be not knowing that though! lol
http://jwbales.us/precal/part6/part6.3.html this is more my thinking
sin(z) = bi/|z|
when bi=0; sin(z) = 0
Hey...X( X( I know what that is X( is just that i thought integers in english was Z as in my place is...
I saw one textbook have it as J and I was like... whats that? lol
Why can't they keep it to just one universal symbol~!
37 mins ago I've written:''Idk how u may call them...Sorry...'' So it's not my fault..:(:(:(
ho ho ho
i blame the chinese... cant trust anyone theat squints all the time..shifty beady little eyes lol
hahahaha :D :D Lool :D :)
\[z=k\pi, k\in Z\]
what is with the latex???
O.o
its new since they upgraded today; nothing works lol,
@satellite73, are u confident with that answer? which method u're using?

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