1. ## e^(iz)

hello...

how would u prove e^(iz) is equla to cosz+isinz???

2. Originally Posted by mpl06c
hello...

how would u prove e^(iz) is equla to cosz+isinz???
Using the "power series" representations of the 3 functions
and seperating the real and imaginary parts of $\displaystyle e^{iz}$.

3. yeah... this is only true when z is real though..
so do i set z:R

4. Originally Posted by mpl06c
yeah... this is only true when z is real though..
so do i set z:R
From the expansion for $\displaystyle e^x$, place x=z to get the expansion for z, and x=iz for the expansion of $\displaystyle e^{iz}$

the exponent of e doesn't have to be real.

5. Originally Posted by mpl06c
hello...

how would u prove e^(iz) is equla to cosz+isinz???
Let $\displaystyle z = \cos{x} + i\sin{x}$.

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

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

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

$\displaystyle = iz$.

Now since $\displaystyle \frac{dz}{dx} = iz$

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

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

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

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

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

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

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

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

$\displaystyle z = Ae^{ix}$, where $\displaystyle A = \pm e^C$.

Now, let's get evaluate the constant. You should know that when $\displaystyle x = 0$ then $\displaystyle z = \cos{0} + i\sin{0} = 1$.

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

$\displaystyle 1 = Ae^0$

$\displaystyle A = 1$.

So we have $\displaystyle z = \cos{x} + i\,\sin{x} = e^{ix}$.

6. hmm ok...
just have to quick question...

1. how do u find e^i

and 2..
say you have(1+i)^i

i understand it all and it gets broken down to

exp(ilog(1+i))

but why cant it also be written as (e^logi)*(e^(1+i))

7. Originally Posted by mpl06c
hmm ok...
just have to quick question...

1. how do u find e^i

and 2..
say you have(1+i)^i

i understand it all and it gets broken down to

exp(ilog(1+i))

but why cant it also be written as (e^logi)*(e^(1+i))
1. Since $\displaystyle e^{ix} = \cos{x} + i\sin{x}$, what happens when $\displaystyle x = 1$?

2. $\displaystyle (1 + i)^i$.

Write $\displaystyle 1 + i$ in polar form.

$\displaystyle 1 + i = \cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}} = e^{\frac{\pi}{4}i}$.

So $\displaystyle (1 + i)^i = (e^{\frac{\pi}{4}i})^i$

$\displaystyle = e^{\frac{\pi}{4}i^2}$

$\displaystyle = e^{-\frac{\pi}{4}}$

8. Originally Posted by mpl06c
hmm ok...
just have to quick question...

1. how do u find e^i

and 2..
say you have(1+i)^i

i understand it all and it gets broken down to

exp(ilog(1+i))

but why cant it also be written as (e^logi)*(e^(1+i))
Because they are not equal!

$\displaystyle e^{a+b}= e^ae^b$, not $\displaystyle e^{ab}= (e^a)^b= (e^b)^a$.

9. Originally Posted by mpl06c
hello...

how would u prove e^(iz) is equla to cosz+isinz???
$\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4 !}+\frac{x^5}{5!}+....$

$\displaystyle Cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+....$

$\displaystyle Sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}-....$

$\displaystyle e^{iz}=1+iz+\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\f rac{(iz)^4}{4!}+\frac{(iz)^5}{5!}+\frac{(iz)^6}{6! }+\frac{(iz)^7}{7!}...$

$\displaystyle =1+iz-\frac{z^2}{2!}-\frac{iz^3}{3!}+\frac{z^4}{4!}+\frac{iz^5}{5!}-\frac{z^6}{6!}-\frac{iz^7}{7!}+\frac{z^8}{8!}+\frac{iz^9}{9!}-....$

$\displaystyle =\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...\right)+i\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-....\right)$

$\displaystyle =Cosz+iSinz$

,

,

### e^iz

Click on a term to search for related topics.