User blog comment:Sbiis Saibian/Googology101 - Part II/@comment-10429372-20141024190431/@comment-5982810-20141025181242

@LP9

Not silly questions at all. I just figured most of the people here were already familiar enough with the basics of complex analysis that I didn't have to explain too much.

The most important equation for complex exponentiation, and how it is actually defined is nicely contained in the equation e^(i*pi) = -1. From this all other complex exponentiation can be derived.

This not an arbitrary definition. It is based on the definition for powers of e...

e^x = lim(n-->inf) (1+x/n)^n

We can exploit this formula to obtain a result for e^(i*pi) since this would be lim(n-->inf) (1+ (i*pi)/n )^n

Note that the number 1 + ( i*pi)/n is a number lying directly, above "1" on the complex plane. As n-->inf this value will approach "1" but always lie a little above it. In complex multiplication the product of two complex numbers is the product of their magnitudes pointed in the direction of the sum of their angles. So when we consider powers of 1 + (i*pi)/n what we find is that since it's radius is only slightly greater than 1, it's first few powers will also have a radius relatively close to 1. Since it's angle is only slightly greater than 0 radians, we will find that for small powers the angle is likewise small. But what happens when we raise 1+(i*pi)/n to the nth power? The little angles add up and approximate pi radians (they are always a little less), furthermore the radius is always a little more than 1. But as n-->inf we find that the result gets closer and closer to having a radius of 1 (from above) at an angle of pi radians (from below). In other words it is approaching -1.

It turns out that whenever we have e^(i*y) it always has a radius of 1 with an angle equal to y radians. This gives us the very useful formula...

e^(i*y) = cos(y) + i * sin(y)

Furthermore when computing e^(x+i*y) we get e^x*cos(y) + i*e^x*sin(y)

So the real part "x" tells us the magnitude while the imaginary part "y" tells us the direction!

Furthermore since we now have e^(x + i*y) we can now also solve ln(x+i*y). This allows us to say that (a+bi)^(c+di) = e^((c+di)*ln(a+bi)). Hence we can now raise any complex number to any complex number (excluding only the case when the base is 0 and the exponent is equal to or less than 0 ).

That's the basics.