e^(pi*i) and more

[mlewis]

You've probably all seen the Euler theorem e^(pi*i)=-1 Me and one of my algebra II classmates have come up with some derivations of this equation for things like the logarithm of imaginary numbers; does anyone know how to take the log of a complex number, ie log (a+bi)?????? We're stuck here and I can't find a proof....

[Guv]

I do not know how to make Greek letters when posting to the forum, So I will use R and A instead of Rho and Theta, which are normally used for radial distance and angle in polar coordinates.

I will be a little sloppy and not use multiply operator with i (squareroot of minus one). Also, ln is usually used instead of log for natural logarithms.

The mathematics which leads to e^iPi = -1 is derived from complex numbers expressed in Polar coordinates.

R = SquareRoot(x^2 + y^2)
A = ArcTangent(y / x )

x + iy = R*e^iA, which is Polar form.

ln(x + iy) = ln(R*e^iA)

ln(x + iy) = ln(R) + ln(e^iA)

ln(x + iy) = ln(R) + iA

[mlewis]

Sweet. Thanks man.