Euler's formula

  • #1
1,197
1
I am to find the imaginary part, real part, square, reciprocal, and absolut value of the complex function:

[tex]y(x,t)=ie^{i(kx-\omega t)} [/tex]
[tex]y(x,t)=i\left( cos(kx- \omega t)+ i sin(kx- \omega t) \right)[/tex]
[tex]y(x,t)=icos(kx- \omega t)-sin(kx- \omega t)[/tex]

the imaginary part is [tex]cos(kx- \omega t)[/tex]

the real part is [tex]-sin(kx- \omega t)[/tex]

the square is:
[tex]-cos^2(kx- \omega t)-2icos(kx- \omega t)sin(kx- \omega t)+sin^2(kx- \omega t)[/tex]
[tex]=-cos^2(kx- \omega t)-isin(2kx-2 \omega t)+sin^2(kx- \omega t)[/tex]
[tex]=-\frac{1}{2}(1+cos(2(kx- \omega t))-isin(2((kx- \omega t))+\frac{1}{2}(1-cos(2(kx- \omega t))[/tex]
[tex]=-cos(2(kx- \omega t))-isin(2((kx- \omega t))[/tex]

the reciprocal is:
[tex]\frac{1}{ie^{i(kx- \omega t)}}[/tex]
[tex]=-ie^{-i(kx- \omega t)}[/tex]
[tex]=-icos(kx- \omega t)-sin(kx- \omega t)[/tex]

absolute value: (not to sure about this...)
[tex]=|icos(kx- \omega t)-sin(kx- \omega t)|[/tex]
[tex]cos^2(kx- \omega t)+sin^2(kx- \omega t)[/tex]
=1???

do these look okay?
 
Last edited:

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
4,780
12
We usually talk about the norm of a complex number, not its absolute value.

You're only missing a square root: |x+iy| = [itex]\sqrt{x^2+y^2}[/itex], but it does not affect the answer.

The rest looks good to me.
 

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