# Homework Help: Euler's formula

1. Feb 7, 2006

### UrbanXrisis

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

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

the imaginary part is $$cos(kx- \omega t)$$

the real part is $$-sin(kx- \omega t)$$

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

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

$$=|icos(kx- \omega t)-sin(kx- \omega t)|$$
$$cos^2(kx- \omega t)+sin^2(kx- \omega t)$$
=1???

do these look okay?

Last edited: Feb 7, 2006
2. Feb 8, 2006

### quasar987

We usually talk about the norm of a complex number, not its absolute value.

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

The rest looks good to me.