View Full Version : Euler's Formula - Need help!
AndersHermansson
Dec24-03, 05:08 PM
I just started studying complex numbers. It says complex numbers in polar form can be expressed as a power of e:
e^ix = cosx + isinx
I don't quite understand how this equation works.
How do i evaluate e^ix? And how does taking e to the power of ix get me a complex number a + bi or even in polar form?
To get all complex numbers, you need the somewhat more general form:
re^{i\theta}=r\sin{\theta}+ir\cos{\theta}
Now, the other direction is also not bad
z=x+iy=\sqrt{x^2+y^2} e^{i\arctan{\frac{x}{y}}}
If you think of r as the radius, and \theta as the angle anti-clockwise from the postive axis, you'll find that this corresponds to switching between polar and rectangular coordinates.
AndersHermansson
Dec24-03, 05:27 PM
So re^{i\theta} is merely a convenience in writing?
mathman
Dec24-03, 05:32 PM
It is more than a convenience. One way to look at it is by power series of eix and compare to power series for sinx and cos x. You will see that eix=cosx+isinx.
AndersHermansson
Dec24-03, 05:35 PM
Ok, yes i think i'm beginning to understand. How do one evaluate xi? For example, what's 32i?
If you are thinking of x^n being the product of n x's, and want to apply that here, you are out of luck. When the exponent is imaginary, it doesn't really carry over. Think of it as a generalization.
Are you familiar with Taylor series?
e^x=1+x+{x^2\over 2}+{x^3\over 3!}+{x^4\over 4!}+...
\sin x=x-{x^3\over 3!}+{x^5\over 5!}-...
\cos x=1-{x^2\over 2}+{x^4\over 4!}-...
The result follows from these.
Well, you can do the following:
3^{2i}=e^{\ln{3} *2i}=\cos{2\ln{3}}+i\sin{2\ln{3}}
If you're feeling a bit more adventerous, there is a power series for 3^x which is going to give you the same result.
AndersHermansson
Dec24-03, 05:50 PM
Ok, thanks! So the only way of evaluating xiy is by using Euler's formula, which we know to be true? I can see the use of the formula more clearly now.
For the record, x^y is usually a multivalued function.
Originally posted by AndersHermansson
Ok, thanks! So the only way of evaluating xiy is by using Euler's formula, which we know to be true? I can see the use of the formula more clearly now.
Think of it as a way to calculate x^{iy}.
phoenixthoth
Dec25-03, 01:48 AM
Originally posted by NateTG
To get all complex numbers, you need the somewhat more general form:
re^{i\theta}=r\sin{\theta}+ir\cos{\theta}
Now, the other direction is also not bad
z=x+iy=\sqrt{x^2+y^2} e^{i\arctan{\frac{x}{y}}}
If you think of r as the radius, and \theta as the angle anti-clockwise from the postive axis, you'll find that this corresponds to switching between polar and rectangular coordinates.
should that be arctan(y/x)?
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