MHB Is $f(z) = \sin(z)/z$ an analytic function on the complex plane?

[Amer]

Find the integral
$\displaystyle \int_C \dfrac{\sin(z)}{z} dz $ where $c: |z| = 1 $

Can I use Cauchy integral formula since sin(z) is analytic

$\displaystyle\int_C \dfrac{\sin(z)}{z} dz = Res(f,0) = 2\pi i \sin(0) = 0$

I tired to compute it without using the formula
$z(t) = e^{it} , 0\leq t\leq 2\pi $

$\displaystyle\int_C \dfrac{\sin(z)}{z} dz = \int_0^{2\pi} \dfrac{\sin(e^{it})}{e^{it}} \cdot i e^{it} dt = i \int_0^{2\pi} \sin(e^{it}) dt $
since it is an odd function on a symmetric interval is zero

So we can treat the integral of a function with removable singularity as analytic functions over closed curve, Cauchy theorem holds for functions with removable singularity..
Is that true Thanks

 

[chisigma]

Find the integral
$\displaystyle \int_C \dfrac{\sin(z)}{z} dz $ where $c: |z| = 1 $

Can I use Cauchy integral formula since sin(z) is analytic

$\displaystyle\int_C \dfrac{\sin(z)}{z} dz = Res(f,0) = 2\pi i \sin(0) = 0$

I tired to compute it without using the formula
$z(t) = e^{it} , 0\leq t\leq 2\pi $

$\displaystyle\int_C \dfrac{\sin(z)}{z} dz = \int_0^{2\pi} \dfrac{\sin(e^{it})}{e^{it}} \cdot i e^{it} dt = i \int_0^{2\pi} \sin(e^{it}) dt $
since it is an odd function on a symmetric interval is zero

So we can treat the integral of a function with removable singularity as analytic functions over closed curve, Cauchy theorem holds for functions with removable singularity..
Is that true Thanks

In fact the function $\displaystyle f(z) = \frac{\sin z}{z}$ is an entire function, i.e. it is analytic in the whole complex plane and the demonstration of that is that it can be represented as Weierstrass infinite product...

$\displaystyle \frac{\sin z}{z} = \prod_{n=1}^{\infty} \{1 - \frac{z^{2}}{(n\ \pi)^{2}}\}\ (1)$

In my opinion a lot of confusion should be removed simply removing the concept of 'removable singularity' from the complex analysis...

Kind regards

$\chi$ $\sigma$

 

[Amer]

In fact the function $\displaystyle f(z) = \frac{\sin z}{z}$ is an entire function, i.e. it is analytic in the whole complex plane and the demonstration of that is that it can be represented as Weierstrass infinite product...

$\displaystyle \frac{\sin z}{z} = \prod_{n=1}^{\infty} \{1 - \frac{z^{2}}{(n\ \pi)^{2}}\}\ (1)$

In my opinion a lot of confusion should be removed simply removing the concept of 'removable singularity' from the complex analysis...

Kind regards

$\chi$ $\sigma$

Is not $\dfrac{\sin(z)}{z}$ undefined at 0 ?
how could it be an entire function ?

Thanks

 

[chisigma]

Is not $\dfrac{\sin(z)}{z}$ undefined at 0 ?
how could it be an entire function ?

Thanks

Writing again the Weierstrass product...

$\displaystyle \frac{\sin z}{z} = \prod_{n=1}^{\infty} \{1 - \frac{z^{2}}{(n\ \pi)^{2}}\}\ (1)$

... You can observe that, setting z=0 in (1), You obtain that for z = 0 is $\displaystyle \frac{\sin z}{z} = 1$...

Kind regards

$\chi$ $\sigma$

 

[Amer]

thanks

 

[alyafey22]

Or you can use the series representation of $$f(z) = \frac{\sin (z) }{z}$$.

 

[Amer]

Or you can use the series representation of $$f(z) = \frac{\sin (z) }{z}$$.

Thanks, I think it is the same as in real integration if f(x) = g(x) Almost everywhere then
int f(x) = int g(x)

 

[ThePerfectHacker]

Define the function $f(z) = \sin(z)/z$ as a function $f:\mathbb{C}^{\times}\to \mathbb{C}$. If you now define $f(0) = 0$ then $f$ is differenciable at $0$. To see this write, for $z\not = 0$,
$$ \frac{f(z)-f(0)}{z-0} = \frac{\sin z}{z^2} $$
The limit of this as $z\to 0$ exists and is equal to $0$. Thus, by definition, $f$ is differenciable at $0$.