Find all solutions of sin(z)=0?

- anonymous

Find all solutions of sin(z)=0?

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- amistre64

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

- anonymous

z= complex

- amistre64

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

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## More answers

- amistre64

thought I erased that odd part lol

- angela210793

z=2*k*pi

- anonymous

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

- angela210793

\[k \in Z\]

- amistre64

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

- angela210793

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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..

- amistre64

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

- anonymous

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

- angela210793

\[R \subset complex no.s\]

- amistre64

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...

- amistre64

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

- anonymous

But only integer which is of the form npi is 0

- anonymous

So the answer should be just one

- anonymous

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

- amistre64

##### 1 Attachment

- amistre64

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

- angela210793

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

- amistre64

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

- amistre64

as opposed to: point = (x,y)

- anonymous

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?

- amistre64

exactly :)

- amistre64

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

- amistre64

every point along the real number line is valid

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

sin(1+0i) = 0?

- amistre64

sin(z) = 0 means at least:
sin(a + bi) = 0

- angela210793

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

- amistre64

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

- anonymous

But polar form just gives you a way to express

- anonymous

Not a complex plane

- amistre64

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

- anonymous

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

- amistre64

it wold resemble the polar form yes...

- angela210793

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

- angela210793

than*

- anonymous

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

- amistre64

i havent blocked them lol

- amistre64

wolfram is good but still has issues with interpreting data

- angela210793

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

- anonymous

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

- angela210793

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

- amistre64

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

- angela210793

K belongs to Z

- anonymous

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

- amistre64

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

- angela210793

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

- amistre64

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

- angela210793

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

- angela210793

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

- amistre64

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

- amistre64

its like using x for a real variable;
they use z for the complex variable

- angela210793

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

- amistre64

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

- anonymous

Integers!

- amistre64

your Z would indicate "any integer" ;

- angela210793

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

- angela210793

Thanks :):):) ^_^

- anonymous

Using 0 = (e^(iz) - e^(-iz)) / 2i
I get stucked when
2iz= log 1
what's next...

- anonymous

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

- amistre64

http://jwbales.us/precal/part6/part6.3.html
this is more my thinking

- amistre64

sin(z) = bi/|z|

- amistre64

when bi=0; sin(z) = 0

- angela210793

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

- amistre64

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

- anonymous

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

- angela210793

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

- anonymous

ho ho ho

- amistre64

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

- angela210793

hahahaha :D :D Lool :D :)

- anonymous

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

- anonymous

what is with the latex???

- angela210793

O.o

- amistre64

its new since they upgraded today; nothing works lol,

- anonymous

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

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