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anonymous
 5 years ago
Find all solutions of sin(z)=0?
anonymous
 5 years ago
Find all solutions of sin(z)=0?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1all of them? and is z the complex variable or just some random variable that is real?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1every odd iteration of pi makes sin(x) = 0 ..... but Z eh

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1thought I erased that odd part lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0z=2kpi? Where K is a real no.? :D

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1I dont know of any sin(z) =0 that exists in the complex part

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Idk how u may call them...Sorry...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if we start out as the [0,inf) real line as our initial ray, then any tine a + 0i occurs we get sin=0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1so at 1+0i and 1+0i times whatever scalar you wanna used for the unit circle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But a+0i isn't a complex number, So you're saying every integer is a complex number, aren't you?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1every 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..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the complex plane includes the real number line; so yes... all real numbers are a subset of the complex plane

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, if so, then the answer is just one, which is 0.

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0\[R \subset complex no.s\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its 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...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when Z = any complex number of the form: a+0i; we get sin(z) = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But only integer which is of the form npi is 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the answer should be just one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But now what confuses me is, x1=0 has a complex solution?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1Z is not retricted to "off of the real number line" so just sating '1' is useless

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Not 1....4 example...k=1> Z=2(1)*pi=2pi=angle 0 k=0>Z=angle 0 k=1>=2*1*pi=angle 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the coordinate system is like the reals; but you define Z as: z = (x + yi)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1as opposed to: point = (x,y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1but to say npi means to say only those z values that are a real multiple of pi is wrong

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1every point along the real number line is valid

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi is a number of the real number line; its a measure of the diatmer of a circle to its circumference

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is the polar form correct, bcuz you are saying sin(1) = 0?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sin(1 +0i) = 0 at least :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sin(z) = 0 means at least: sin(a + bi) = 0

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0sin(1)=0....? how come?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sin(a + 0i) = 0 for any given a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But polar form just gives you a way to express

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1polar form can be used to interpret the complex plane into a real plane;

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm., check this out http://www.wolframalpha.com/input/?i=sin%281%2B0i%29

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it wold resemble the polar form yes...

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Don'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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't you think, we are leaving out someone in this discussion? The asker!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i havent blocked them lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1wolfram is good but still has issues with interpreting data

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0haha lol :D.....maybe he has got it...:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey..i'm still here,get confuse @_@

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Oops....@Amistre do u agree with me that the answer is Z=2kpi?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1im open to me being wrong :) but I would have to see how sin(z) is sposed to be interpreted :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm reading this post, try to digest, what do u guys think: http://answers.yahoo.com/question/index?qid=20080529224140AAGdnOg

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1angela, I would have to research it to tell :)

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Ok.....I have always thought math is same everywhere...but...o.O

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1maybe sin(z) = 0 where e^(ix) = e^(ix) :)

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0maybe....ur the smart guy here anyway :P

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0u have Z={..........3,2,1,0,1,2,3.........} right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1no; z is beig used in this context to indicate the complex variable and not the set of integers

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its like using x for a real variable; they use z for the complex variable

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Really??????? well in my answer i meant for this kind of Z.....and wht do u call ''my Z''?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1but all signs point to the answer being npi where n is any integer

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1your Z would indicate "any integer" ;

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0ahaa...that's wht i meant....:S:S

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Using 0 = (e^(iz)  e^(iz)) / 2i I get stucked when 2iz= log 1 what's next...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@angela210793: You never seemed to be not knowing that though! lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1http://jwbales.us/precal/part6/part6.3.html this is more my thinking

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when bi=0; sin(z) = 0

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Hey...X( X( I know what that is X( is just that i thought integers in english was Z as in my place is...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1I saw one textbook have it as J and I was like... whats that? lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why can't they keep it to just one universal symbol~!

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.037 mins ago I've written:''Idk how u may call them...Sorry...'' So it's not my fault..:(:(:(

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i blame the chinese... cant trust anyone theat squints all the time..shifty beady little eyes lol

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0hahahaha :D :D Lool :D :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is with the latex???

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its new since they upgraded today; nothing works lol,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@satellite73, are u confident with that answer? which method u're using?
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