Problem while playing with Bessel functions

[saybrook1]

Homework Statement

I have run into a number of problems while working through problems regarding Bessel and Modified Bessel Functions. At one point I run into $i^{m}e^{\frac{im\pi}{2}}$ and it needs to equal $(-1)^m$ but I'm not sure how it does. This came up while trying to solve an identity for a modified Bessel function in problem 14.5.2 in Arfken and Weber 'Mathematical Methods for Physicists'. Any help would be greatly appreciated, thank you.

Homework Equations

$i^{m}e^{\frac{im\pi}{2}}$=$(-1)^m$ but how??

The Attempt at a Solution

I solved for this by expanding the exponent into cos and sin but I wind up with $i^{m} [cos\frac{m\pi}{2}+(-1)^{m}\sin\frac{m\pi}{2}]$ and don't know how that can equal $(-1)^m$

 

[Samy_A]

Homework Statement

I have run into a number of problems while working through problems regarding Bessel and Modified Bessel Functions. At one point I run into $i^{m}e^{\frac{im\pi}{2}}$ and it needs to equal $(-1)^m$ but I'm not sure how it does. This came up while trying to solve an identity for a modified Bessel function in problem 14.5.2 in Arfken and Weber 'Mathematical Methods for Physicists'. Any help would be greatly appreciated, thank you.

Homework Equations

$i^{m}e^{\frac{im\pi}{2}}$=$(-1)^m$ but how??

The Attempt at a Solution

I solved for this by expanding the exponent into cos and sin but I wind up with $i^{m} [cos\frac{m\pi}{2}+(-1)^{m}\sin\frac{m\pi}{2}]$ and don't know how that can equal $(-1)^m$

There is an error in $i^{m} [cos\frac{m\pi}{2}+(-1)^{m}\sin\frac{m\pi}{2}]$, that should be $i^{m} [cos\frac{m\pi}{2}+i\sin\frac{m\pi}{2}]$

An easy way to solve your exercise it to consider four case separately: $m=0, m=1,m=2, m=3$.

 

[saybrook1]

There is an error in $i^{m} [cos\frac{m\pi}{2}+(-1)^{m}\sin\frac{m\pi}{2}]$, that should be $i^{m} [cos\frac{m\pi}{2}+i\sin\frac{m\pi}{2}]$

An easy way to solve your exercise it to consider four case separately: $m=0, m=1,m=2, m=3$.

You're right, I just typed it out incorrectly. Okay, I'll try and consider those cases; Thank you.

 

[SteamKing]

Homework Statement

I have run into a number of problems while working through problems regarding Bessel and Modified Bessel Functions. At one point I run into $i^{m}e^{\frac{im\pi}{2}}$ and it needs to equal $(-1)^m$ but I'm not sure how it does. This came up while trying to solve an identity for a modified Bessel function in problem 14.5.2 in Arfken and Weber 'Mathematical Methods for Physicists'. Any help would be greatly appreciated, thank you.

Homework Equations

$i^{m}e^{\frac{im\pi}{2}}$=$(-1)^m$ but how??

The Attempt at a Solution

I solved for this by expanding the exponent into cos and sin but I wind up with $i^{m} [cos\frac{m\pi}{2}+(-1)^{m}\sin\frac{m\pi}{2}]$ and don't know how that can equal $(-1)^m$

If you express i in polar form, you have i = e^(iπ/2). From there, you should be able to find i^m in polar form quite easily and then to find the product of i^m and e^(imπ/2) in polar form.

Soon after that, a vist from your Uncle Bob can be expected.