# Thread: exp(iat+b) = cos(at+b) + isin(at+b) ???

1. ## exp(iat+b) = cos(at+b) + isin(at+b) ???

exp(iat+b) = cos(at+b) + isin(at+b)

pls i need 2 know

2. Do you mean

Prove $\displaystyle e^{i(at+b)}= \cos{(at+b)}+i\sin(at+b)$

??

3. OH its i(at+b) kk thx man

4. To make the question easier make $\displaystyle x=at+b$

And let $\displaystyle z = \cos{x} + i\sin{x}$.

It is also important to note that when $\displaystyle x = 0, z = 1$.

Then

$\displaystyle \frac{dz}{dx} = -\sin{x} + i\cos{x}$

$\displaystyle = i^2\sin{x} + i\cos{x}$

$\displaystyle = i(\cos{x} + i\sin{x})$

$\displaystyle = iz$.

So $\displaystyle \frac{dz}{dx} = iz$.

Rearranging and integrating gives

$\displaystyle \frac{1}{z}\,\frac{dz}{dx} = i$

$\displaystyle \int{\frac{1}{z}\,\frac{dz}{dx}\,dx} = \int{i\,dx}$

$\displaystyle \int{\frac{1}{z}\,dz} = ix + C_1$

$\displaystyle \ln{|z|} + C_2 = ix + C_1$

$\displaystyle \ln{|z|} = ix + C$, where $\displaystyle C = C_1 - C_2$

$\displaystyle |z| = e^{ix + C}$

$\displaystyle |z| = e^Ce^{ix}$

$\displaystyle z = \pm e^C e^{ix}$

$\displaystyle z = Ae^{ix}$.

Now to find the constant, recall that when $\displaystyle x = 0, z = 1$.

So

$\displaystyle 1 = Ae^{0i}$

$\displaystyle A = 1$.

Therefore

$\displaystyle z = e^{ix}$.

Thus this proves Euler's Formula:

$\displaystyle z = \cos{x} + i\sin{x} = e^{ix} \Rightarrow \cos{at+b} + i\sin{at+b} = e^{i(at+b)}$

*Thanks to Prove it for the code