hello...

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

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- Feb 11th 2010, 06:21 PM #1

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- Feb 11th 2010, 06:53 PM #2

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- Feb 11th 2010, 07:09 PM #3

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- Feb 11th 2010, 07:15 PM #4

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- Feb 11th 2010, 07:18 PM #5
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 $\displaystyle z = \cos{x} + i\,\sin{x} = e^{ix}$.

- Feb 11th 2010, 07:48 PM #6

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- Feb 12th 2010, 01:57 AM #7
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 $\displaystyle (1 + i)^i = (e^{\frac{\pi}{4}i})^i$

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

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

- Feb 12th 2010, 02:01 AM #8

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- Feb 12th 2010, 04:24 AM #9

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

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

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

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

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

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

$\displaystyle =Cosz+iSinz$

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