1

Using Euler's formula prove the following: $$\sin(z)=0\Leftrightarrow z=\pi k\text{ for } k\in\mathbb{Z}.$$

I don't even know where to start.

Robert Z
  • 145,942
Joanna M
  • 75
  • 1
    Have you tried using Euler's formula? – Arthur Feb 02 '18 at 09:53
  • Can we use https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi ? – lab bhattacharjee Feb 02 '18 at 09:54
  • 4
    You could start by applying Euler's formula, but hey, that's just me. – 5xum Feb 02 '18 at 09:55
  • The key for that task is to prove $\sin(z)=\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$. You see now, that all the roots of $\sin(z)$ are the roots of $\sin(x)$. – Fakemistake Feb 02 '18 at 10:09
  • A good place to start is the definition of $\sin(z)$ when $z$ is a complex number. – Per Erik Manne Feb 02 '18 at 11:21
  • Any Question that asks about "if and only if" can be started by looking at the two directions separately and asking if one is easy or easier than the other. In this case you should notice that one of the directions is a very familiar statement about trigonometry for real numbers. – hardmath Feb 03 '18 at 17:25

3 Answers3

4

Hint: Note that the sine with complex argument is defined as $$\sin(z):=\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}= \frac{e^{-y}e^{ix}-e^{y}e^{-ix}}{2i}$$ where $z=x+iy$ with $x,y\in\mathbb{R}$. Then by using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$? we get $$\sin(z)=0\Leftrightarrow e^{-y}(\cos(x)+i\sin(x))-e^{y}(\cos(x)-i\sin(x)) \Leftrightarrow \begin{cases} \sin(x)(e^y+e^{-y})=0 \\ \cos(x)(e^y-e^{-y})=0 \end{cases}$$ Can you take it from here?

John
  • 3
  • 4
Robert Z
  • 145,942
2

It depends on want you want to prove and how you define things... A way to see this question is to ask if $$ \forall z\in \mathbb{C}, \quad \sin(z)=0 \qquad \Longleftrightarrow \qquad z=\pi k , \quad k\in \mathbb{z}$$ knowing that the same statement is true over $\mathbb{R}$... With $z=x+iy$ \begin{align*} \sin(z):= \frac{e^{iz}-e^{-iz}}{2i} =0 &\qquad \Longleftrightarrow \qquad e^{iz}=e^{-iz} \\ &\qquad \Longleftrightarrow \qquad e^{ix}e^{-y}=e^{-ix}e^{y} \end{align*} 1) $e^{ix}e^{-y}=e^{-ix}e^{y} \Rightarrow |e^{ix}e^{-y}|=|e^{-ix}e^{y}| \Rightarrow e^{-y}=e^{y} \Rightarrow y=0$, since $t\mapsto e^t$ is strictly increasing.

2)With $y=0$,
\begin{align*} e^{ix}e^{-y}=e^{-ix}e^{y} &\quad \Rightarrow \quad e^{ix} =e^{-ix} \\ &\quad \Rightarrow \quad \cos(x)+i\sin(x) =\cos(-x)+i\sin(-x) \\ &\quad \Rightarrow \quad \cos(x)+i\sin(x) =\cos(x)-i\sin(-x) \\ &\quad \Rightarrow \quad \sin(x) =0 \\ &\quad \Rightarrow \quad x=k\pi, \quad k\in \mathbb{Z} \\ \end{align*} So $z=k\pi+i*0=k\pi$, the converse being obvious.

Netchaiev
  • 4,811
1

Euler's formula says that the complex exponential $e^{iz}$ can be expressed as a sum of sinusoidals: $$e^{iz}=cos(z) + i\ sin(z)$$. Does that help?

Ranjeev Grewal
  • 112