Find Roots of sin z: Solutions & Explanations

[ognik]

Looking for someone to check my working & answers please. Problem is 'find all the zeros of sin z'

A) sin z = sin(x+iy) = sin(x)cosh(y) + i cos(x)sinh(y)
Roots are when sin(x)cosh(y) = 0 = cos(x)sinh(y)

$If \: sinh(y)=0, then \: cosh(y)=1 \: (cosh^2 - sinh^2=1) $
$ \therefore sin(x) = 0, \therefore x = n\pi, n=0,\pm1,\pm2 ... $
$ y = arcsinh(0) = 0 $
$ \therefore $ 1st root is $ (0,n\pi) $

B) I also would like to do it using exponents:
$ sin z = \frac{1}{2i}\left(e^{z}-e^{-z}\right) $
$ \therefore \left(e^{z}-e^{-z}\right) = 0 $
$ \therefore e^{z} = e^{-z}, \: \therefore x+iy = -(x+iy)$
This just gives x = 0 = y which seems trivial?
Thanks in advance.

 

[I like Serena]

Looking for someone to check my working & answers please. Problem is 'find all the zeros of sin z'

A) sin z = sin(x+iy) = sin(x)cosh(y) + i cos(x)sinh(y)
Roots are when sin(x)cosh(y) = 0 = cos(x)sinh(y)

$If \: sinh(y)=0, then \: cosh(y)=1 \: (cosh^2 - sinh^2=1) $
$ \therefore sin(x) = 0, \therefore x = n\pi, n=0,\pm1,\pm2 ... $
$ y = arcsinh(0) = 0 $
$ \therefore $ 1st root is $ (0,n\pi) $

B) I also would like to do it using exponents:
$ sin z = \frac{1}{2i}\left(e^{z}-e^{-z}\right) $
$ \therefore \left(e^{z}-e^{-z}\right) = 0 $
$ \therefore e^{z} = e^{-z}, \: \therefore x+iy = -(x+iy)$
This just gives x = 0 = y which seems trivial?
Thanks in advance.

Hi ognik,

(A) is fine, although saying the 1st root is $ (0,n\pi) $ isn't proper.
It should say something like "the roots are $n\pi$ where $n$ is an integer".
That's because it's not only the 1st root, but all roots.
And $ (0,n\pi) $ either denotes the open interval from $0$ to $n\pi$, or the point with x-coordinate $0$ and y-coordinate $n\pi$.
Neither is a root.I'm afraid that in (B) the formula for $\sin z$ is wrong.
It should be $ \sin z = \frac{1}{2i}\left(e^{iz}-e^{-iz}\right) $

Continuing, we get:
$$ e^{i(x+iy)} = e^{-i(x+iy)} $$
Now, when dealing with complex numbers, we can't just take the natural logarithm, because it is a multivalued function.
Properly, we need to account for the fact that the function $e^{ix}$ has a period of $2\pi$, which results in:
$$ i(x+iy) = -i(x+iy + 2\pi n)
\Rightarrow x+iy = -x -iy - 2\pi n
\Rightarrow 2(x + \pi n) + 2iy = 0
\Rightarrow x = -\pi n \wedge y = 0
$$

 

[ognik]

Thanks ILS, again a careless mistake :-(

I am still coming to terms with some of the variable meanings in math, I had intended $ (n \pi, 0) $ to represent the values (x, y), not an interval or coordinate; so should I rather have said the roots are $ n \pi + i0, n=0, \pm1,... $?
Then - if we were talking about the roots of a multi-varied (say x,y,z), real equation - how would I properly write the roots?

 

[I like Serena]

Thanks ILS, again a careless mistake :-(

I am still coming to terms with some of the variable meanings in math, I had intended $ (n \pi, 0) $ to represent the values (x, y), not an interval or coordinate; so should I rather have said the roots are $ n \pi + 0i, n=0, \pm1,... $?
Then - if we were talking about the roots of a multi-varied (say x,y,z), real equation - how would I properly write the roots?

In an equation with real $x,y,z$, we have roots that are elements of $\mathbb R^3$. Those elements are represented with $(x,y,z)$.
In an equation with complex numbers $z \in \mathbb C$ the roots are indeed like $ n \pi + i0$ or just $n\pi$.
In an equation with real $x,y$, we can have roots like $(n\pi, 0)$, which are elements of $\mathbb R^2$.

So your equation $\sin z = 0$ has roots $n\pi$, while the equation $\sin(x+iy)=0$ has roots $(n\pi, 0)$.