Complex Arguments!

I guess there is no need of going into the complex numbers.

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We need to find the solutions of $\sin(x+yi)-\sin(x-yi)=0$ .

Since $\sin(x+yi)-\sin(x-yi)=2\cos(x)\sin(yi)$ , we must have $\cos x=0$ for it to be true for any $y$ .

We know that $\cos(x)=0$ when $x=\dfrac{n\pi}{2}$ where $n=1,3,5,\cdots,1285$ because $\dfrac{n\pi}{2}<2020$ .

This makes a total of $\boxed{643}$ solutions.

$\text{sin}(x + yi) = \text{sin }x\text{ cosh }y + i\text{ cos }x\text{ sinh }y$

$\text{sin}(x - yi) = \text{sin }x\text{ cosh}(-y) + i\text{ cos }x\text{ sinh}(-y)$

$\hspace{50pt} = \text{sin }x\text{ cosh }y - i\text{ cos }x\text{ sinh }y$

Equating both the expression we get,

$\text{sin }x\text{ cosh }y + i\text{ cos }x\text{ sinh }y = \text{sin }x\text{ cosh }y - i\text{ cos }x\text{ sinh }y$

$\Rightarrow 2i \text{cos }x\text{ sinh }y = 0$

For above equation to be true for any $y$ , $\text{ cos }x$ must be $0$ .

$\text{cos }x = 0$

$\Rightarrow \displaystyle x = \frac{(2n - 1)\pi}{2}\;n \geq 1$

$\displaystyle 0 < x < 2020$

$\Rightarrow\displaystyle 0 < \frac{(2n - 1)\pi}{2} < 2020$

$\Rightarrow\displaystyle 0 < 2n - 1 < \frac{4040}{\pi}$

$\Rightarrow\displaystyle 1 < 2n < \frac{4040}{\pi} + 1$

$\Rightarrow\displaystyle \frac{1}{2} < n < \frac{\frac{4040}{\pi} + 1}{2}$

$\Rightarrow 0.5 < n < 643.4859$

$\Rightarrow 1 \leq n \leq 643$

So, there are 643 possible values of $x$ .

We assume that $\sin(x+yi)=\sin(x-yi).\quad\quad (*)$ When the argument of the sine is a complex number $z,$ then $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}.$ Then the equation $(*)$ can be express as $\frac{e^{i(x+yi)}-e^{-i(x+iy)}}{2i}=\frac{e^{i(x-yi)}-e^{-i(x-iy)}}{2i}.$ Multiplying both sides of this equation by $2i,$ simplifying the exponents and moving all the terms to the left side, we get $e^{xi-y}-e^{y-xi}-e^{xi+y}+e^{-xi-y}=0.$ Combining the first term with the third and the second with the fourth, we obtain $e^{xi}(e^{-y}-e^{y})-e^{-xi}(e^y-e^{-y})=0.$ That can be factor to $(e^{xi}+e^{-xi})(e^{-y}-e^{y})=0.$ Replacing $y$ by 1 and noticing that $e^{-1}-e\neq 0,$ it follows that $e^{xi}+e^{-xi}=0.$ Since $\cos x = \frac{e^{xi}+e^{-xi}}{2},$ then the previous equation implies that $\cos x =0.$ Therefore, $x=\frac{\pi}{2}+n\pi,$ where the $n$ is any integer number. Now, to find the number of possible values of $x$ in the interval $(0, 2020),$ we need to solve the inequality $0<\frac{\pi}{2}+n\pi<2020.$ Solving the inequality with respect to $n,$ it results that $-.5< n<642.486.$ Then $n$ can take 643 different values and, therefore, there are $\boxed{643}$ possible values for $x$ in the given interval.

Another alternate explanation:

$sin(A) = sin(B) \iff A = n\pi + (-1)^nB$ for some integer $n$ .

Therefore, $x + iy = n\pi + (-1)^n(x - iy)$ .

If $n$ is even, then $x + iy = n\pi + x - iy \implies 2iy = n\pi\ \implies y = \frac{-in}{2}\pi$ .

If $n$ is odd, then $x + iy = n\pi - x + iy \implies x = n\pi - x \implies x = \frac{n}{2}\pi$ .

(Now not considering $n$ to be specifically odd or even.) Combining the two results, when $y = -in\pi$ , the equation is satisfied by any $x$ . For all other values of $y$ , the equation is satisfied by $x = \frac{2n - 1}{2}\pi$ .

Therefore, to satisfy any real value of $y$ , $x = \frac{2n - 1}{2}\pi$ (which are the solutions to $cos(x) = 0$ ). Taking into account the upper bound on $x$ , we get $n < \frac{4040 + \pi}{2\pi} = 643.485...$ , hence $643$ solutions.