Power Series, Complex Exponentiation

Complex Exponentiation

E^x = 1 + x + x²/2 + x³/6 + x⁴/24 + … + xⁿ/n! + …

This power series converges for any real x, hence E^z is analytic, and its circle of convergence is the entire plane.

Let f be the complex function E^z. We know how f behaves on the real line, so let's stroll up the imaginary axis. Replace x with ix in the above taylor series, and separate the real and imaginary components.

real = 1 - x²/2 + x⁴/24 - x⁶/6! + x⁸/8! - …
imaginary = x - x³/6 + x⁵/5! - x⁷/7! + …

These are the taylor series for cosine and sine respectively. As we move up the imaginary axis, f(z) runs around the unit circle, its real component acting as cosine and its imaginary component acting as sine.

The derivative of f can be computed term by term, and the result is the same taylor series back again, hence f′ = f. This is not a surprise, since we already saw this in real space.

Let t be a real variable, time if you like, and ask how f changes as we move in the "real" direction. The change in f is f′ times the direction of travel, or f′×1 in this case. Yet f′ = f, so the change in f is equal to f. In other words, the change is radial. As t advances, f(z+t) moves away from the origin. The rate of instantaneous change is proportional to the distance, and we've solved this differential equation before. The solution is an exponential curve, multiplied by a constant that is set by the initial conditions. In this case we obtain the following.

f(z+t) = f(z)×E^t

Let z be a point on the imaginary axis, which has an image on the unit circle. In other words, we have selected an angle θ in polar coordinates. Then let t run parallel to the x axis, and f(z+t) runs away from the origin at an exponential rate. If t moves to the left, the image approaches the origin. In polar coordinates, θ is constant, and r = E^t.

Note that f(z) is never equal to 0.

Consider f(z₁+z₂), which we will write as f(a+bi + c+di). By the above, this is equal to f(bi+di)×E^a+c. Remember that f wraps the imaginary axis around the unit circle. In fact the distance up the imaginary axis becomes the angle around the unit circle. In polar coordinates, the angle of b+d is the angle of b plus the angle of d. This is the same as multiplication, according to Demoivre's Formula. Therefore f(bi+di) = f(bi)×f(di). The entire expression can be rewritten this way.

f(bi) × f(di) × E^a × E^c

f(bi)×E^a × f(di)×E^c

f(a+bi) × f(c+di)

f(z₁) × f(z₂)

Complex exponentiation carries addition over to multiplication, just as it did in real space.

Note that f(z)×f(-z) = 1.

Don't assume E^ab is E to the a to the b. This only works for real variables. Let a = 2πi and b = ½. Now E^ab = -1, but E^a is 1, and its square root could be construed as 1, rather than -1.

Trig Functions

cos(z) = ½ ( E^iz + E^-iz )

sin(z) = ½ ( E^iz - E^-iz )

Note the derivative of sine is still cosine, and the derivative of cosine is still -sine.

There's lots more to say about the exponential, log, trig, and hyperbolic trig functions over the complex plane, but this section is suppose to be about power series, so let's get back to it.