# e^(iz)

• Feb 11th 2010, 06:21 PM
mpl06c
e^(iz)
hello...

how would u prove e^(iz) is equla to cosz+isinz???
• Feb 11th 2010, 06:53 PM
Quote:

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 $e^{iz}$.
• Feb 11th 2010, 07:09 PM
mpl06c
yeah... this is only true when z is real though..
so do i set z:R
• Feb 11th 2010, 07:15 PM
Quote:

Originally Posted by mpl06c
yeah... this is only true when z is real though..
so do i set z:R

From the expansion for $e^x$, place x=z to get the expansion for z, and x=iz for the expansion of $e^{iz}$

the exponent of e doesn't have to be real.
• Feb 11th 2010, 07:18 PM
Prove It
Quote:

Originally Posted by mpl06c
hello...

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

Let $z = \cos{x} + i\sin{x}$.

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

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

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

$= iz$.

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

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

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

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

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

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

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

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

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

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

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

So $1 = Ae^{0i}$

$1 = Ae^0$

$A = 1$.

So we have $z = \cos{x} + i\,\sin{x} = e^{ix}$.
• Feb 11th 2010, 07:48 PM
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))
• Feb 12th 2010, 01:57 AM
Prove It
Quote:

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 $e^{ix} = \cos{x} + i\sin{x}$, what happens when $x = 1$?

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

Write $1 + i$ in polar form.

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

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

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

$= e^{-\frac{\pi}{4}}$
• Feb 12th 2010, 02:01 AM
HallsofIvy
Quote:

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!

$e^{a+b}= e^ae^b$, not $e^{ab}= (e^a)^b= (e^b)^a$.
• Feb 12th 2010, 04:24 AM
Quote:

Originally Posted by mpl06c
hello...

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

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4 !}+\frac{x^5}{5!}+....$

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

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

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

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

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

$=Cosz+iSinz$