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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

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...

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?

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

\[R \subset complex no.s\]

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?

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)

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

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...

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...:)

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

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

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?