Zeroes of sin(x) Consider the function f = $\sin(x)$ defined as 
$$ \sin(x) = \frac{e^{ix}- e^{-ix}}{2i} $$
How to prove that the only zeroes of this function lie on the line $i = 0$ in the complex plane and furthermore that those zeroes are of the form $\pi k$ for $k \in \Bbb{Z}$. To begin the first step I note that we really just interested in the zeroes of 
$$  e^{ix}- e^{-ix} $$ 
Furthemore consider a zero x such that $i \ne 0 $ we can declare $x = a + bi$ and therefore 
$$ e^{ai - b} - e^{b - ai} = 0 $$ 
From here it follows that 
$$ ai - b = b - ai$$ 
We note that that the only circumstance this can happen is if $b = ai$ and since b,a are real number it then MUST be the case that $b = 0$ resulting in a contradiction.
For the second phase I note that 
$$ e^{ix}  = e^{-ix} $$ 
Obviously we desire $e^{ix} = \pm 1 $. Given by definition that $e^{i\pi} = -1$ and that integer powers of $-1$ are also in the set $1,-1$ it then follows that each integer times pi is a zero. 
But how do I know that integer multiples of $pi$ are the ONLY values for which $e^{ix} = \pm 1$ ?
 A: We have $\sin(z) = \sin(x+iy) = \frac{e^{ix-y}-e^{-ix+y}}{2i}$
$\frac{e^{ix}e^{-y}-e^{-ix}e^{y}}{2i}$
So $|\sin(z)| = \left|\frac{e^{ix}e^{-y}-e^{-ix}e^{y}}{2i}\right|$ 
$=\frac{\left|e^{ix}e^{-y}-e^{-ix}e^{y}\right|}{2}$
$\geq \left|\frac{\left|e^{ix}e^{-y}\right|-\left|e^{-ix}e^{y}\right|}{2}\right|$
Because $|e^{ix}| = 1$, we can simplify,
$= \left|\frac{e^{-y}-e^{y}}{2}\right|$
so we've found out $|\sin(x+iy)| \geq \left|\frac{e^{-y}-e^{y}}{2}\right|$. But this can only equal 0 if $y=0$, aka $z = x+0y$ for $x \in \mathbb{R}$.
A: $$sin(x)=0$$
Is the same as:
$$\frac{1}{2}ie^{-ix}-\frac{1}{2}ie^{ix}=0$$
$$x=\pi n$$
(with n is the element of Z - the set of integers)
A: From the equation $\mathrm e^{\mathrm ix} = \mathrm e^{-ix}$ we get by multiplying both sides by $\mathrm e^{\mathrm ix}$ the equation $\mathrm e^{\mathrm 2ix}=1$ (that is an equivalence transformation because the exponential function is nowhere $0$).
Now assume that $x_1$ and $x_2$ are zeroes of the $\sin x$. Then $\mathrm e^{2\mathrm ix_1}=1$ and $\mathrm e^{2\mathrm ix_2}=1$ and thus $\mathrm e^{2\mathrm i(x_1-x_2)}=\mathrm e^{2\mathrm ix_1}/\mathrm e^{2\mathrm ix_2}=1$, that is, $x_1-x_2$ is also a zero of $\sin x$.
Now from the defining formula for $\sin x$, it is easy to show that $\sin\overline x = \overline{\sin x}$, therefore if $x$ is a zero, then also $\overline x$ is. But then, from what we just have shown, $x-\overline x$ also is a zero of $\sin x$.
But $x-\overline x = 2\mathrm i\operatorname{Im} x$, and thus $\mathrm e^{2\mathrm i(x-\overline x)} = e^{-4\operatorname{Im} x}$. But that is a real exponent, and thus the only solution of $e^{-4\operatorname{Im} x}=1$ is $\operatorname{Im} x=0$, that is, $x$ must be real.
We can use the power series of the sine function (derived from the power series of the exponential function) to verify that $\sin 1>0$ and $\sin 4<0$, therefore since the exponential function, and therefore the sine function, is continuous, there must be at least one zero in between; that is, there is at least one positive zero.
Since the exponential function is analytic, the sine function is, too, and thus we know that there is a smallest positive zero (because otherwise $0$ would be an accumulation point of zeroes, and thus the sine function would have to be constant, which we know it isn't). Let's call that smallest positive zero $x_0$.
Now we know that $x_0$ being a zero means $\mathrm e^{2\mathrm ix_0}=1$ from which we get $\mathrm e^{2\mathrm inx_0}=1$ for any $n\in\mathbb Z$ by simply taking the appropriate power of the equation. Therefore $n\mathbb x_0$ is also a zero for any $n\in \mathbb Z$.
Now assume $x_1$ is another positive zero of $\sin x$. Then there exists an $r>1$ such that $x_1=rx_0$. But then, $x_1-\lfloor r\rfloor x_0 = \left(r-\lfloor r\rfloor\right)x_0$ would also be a zero, but that would be a positive zero smaller than $x_0$, in contradiction to the assumption that $x_0$ is the smallest positive zero. Since $\sin(-x)=-\sin x$ (easily verified with the defining formula), the same holds also for negative zeroes. Thus all zeroes are of the form $nx_0$, $n\in \mathbb Z$.
The only thing missing is that $x_0=\pi$, where $\pi$ is defined as the ratio of the circumference and diameter of a cycle.
To prove this, let's first define a second function,
$$\cos x = \frac{\mathrm e^{\mathrm ix} + \mathrm e^{-\mathrm ix}}{2}$$
One easily verified that $(\sin x)^2 + (\cos x)^2 = 1$, thus obviously all the points $(\cos x,\sin x)\in\mathbb R^2$ are on the unit circle. However we have to prove that we get the full unit cycle. In particular, we are going to prove that for $x\in[0,2x_0]$ we go around the unit circle exactly once.
First, we notice that $\mathrm e^{\mathrm ix_0}=-1$. This must be the case because the square is $1$ (that gives exactly the condition for $x_0$ being a zero of the sine), but the value itself cannot be $1$ (otherwise $\frac {x_0}{2}$ would be a zero as well, in contradiction to $x_0$ being the smallest positive zero), and the only other value that squares to $1$ is $-1$.
From that, one can easily verify that $\sin(x+x_0)=-\sin x$ and $\cos(x+x_0)=-\cos x$. Also from the fact that the derivative of the exponential function is the exponential function itself (easily verified from the power series) one readily checks that $\frac{\mathrm d}{\mathrm dx}\sin x=\cos x$ and $\frac{\mathrm d}{\mathrm dx}\cos x=-\sin x$.
We also already know that between $0$ and $x_0$ the sine is positive (thanks to continuity it cannot change sign anywhere but on zeroes). Using that fact together with the equation
$$\mathrm e^{\mathrm ix} = \cos x + \mathrm i\sin x$$
(which is easily verified from the definitions of $\sin x$ and $\cos x$) we also get $\mathrm e^{\mathrm i x_0/2}=\mathrm i$ (the square is $-1$ and the imaginary part is positive). With that knowledge we can now also verify that $\cos x = \sin\left(\frac{x_0}{2}-x\right)$. Especially the zeroes of $\cos x$ are at $x_0\left(n+\frac12\right)$.
Now we have all parts together: The curve $x\mapsto(\cos x,\sin x)$ starts at $(1,0)$ for $x=0$. For $0<x<\frac{x_0}{2}$ the sine is positive and strictly monotone growing, while the cosine is positive and strictly monotone falling. At $x=\frac{x_0}{2}$ it reaches the point $(0,1)$. For $\frac{x_0}{2}<x<x_0$ the sine is positive and strictly monotone falling, and the cosine is negative and strictly monotone falling. At $x=x_0$ the point (-1,0)$ is reached. I omit the rest because it should be clear; together with continuity all this gives that we indeed go round the unit circle exactly once, without ever going back on the way.
Now all we have to do is to calculate the length of that curve. The length is obtained by integrating the length of the derivative vector. The derivative vector is $(-\sin x,\cos x)^T$, and its length is $(-\sin x)^2+(\cos x)^2=1$. Thus the circumference of the unit circle is $C=\int_0^{2 x_0} 1\,\mathrm dx = 2x_0$. Obviously the diameter of the unit circle is $D=2$, thus we have $x_0 = C/2 = C/D = \pi$.
