Let a sequence of points approach the limit point c in the complex plane. If infinitely many points in the sequence are zeros, then f(c) = 0 by continuity. The zeros of an analytic function are isolated, so we have a contradiction. The only way out is to set f equal to the 0 function.

Let two analytic functions f and g attain the same values on c, and points arbitrarily close to c. Each has a power series, with a circle of convergence centered at c. Let r be the lesser of the two radii. Thus the power series for f-g converges out to a radius of r. What does this power series look like? Since f and g are equal on the sequence approaching c, f-g is 0 as it approaches c. By the above, f-g = 0, and the series for f and the series for g are the same. This means f = g, at least on the circle centered at c, with radius r.

There is a good chance that convergence stopped at r because of a pole. For example, 1/(1-z) converges up to the unit circle, and no farther, but the function is actually defined on the entire plane, except for the pole at 1. So - f and g are equal on a disk of radius r, but are they equal everywhere, or do they diverge later on?

One could set f = 1 above the real line and f = -1 below the real line. This is analytic wherever it is defined, yet the behavior of f below the real line does not affect or predict the behavior of f above the real line. Disconnected domains can present independent functions. With this in mind, we will try to predict the behavior of f across its connected domain. That's the best we can do.

If you are a topologist, you might wonder about connected verses path connected in the following proof. Well - the domain of an analytic function is always an open set in the plane, hence connected and path conected are the same.

Let d be a point in the same connected component as c, and suppose f(d) ≠ g(d). Draw a path from c to d. Let r(t) be the radius of convergence of the power series of f, or g, whichever is smaller, at a point t along the path. For each t, establish an open disk of radius ½r(t), centered at t. These open sets cover the path. Since the path is compact, a finite subcover will do. This establishes a chain of intersecting open disks, each half as large as it needs to be to assure convergence of f and g.

Draw a finite chain of segments conecting each disk's center to the next. This begins with c and ends with d. How much wiggle room do we have on either side of this chain for convergence? Let s be the smallest radius of the open disks in the chain. If adjacent disks, of size s, barely overlap, convergence extends from their centers out to a distance of 2s, Which is sqrt(3)s from the midpoint of the connecting segment. Along the segment, we can move a distance s on either side, and f and g are analytic. The chain has thickened to a road of width 2s, with a few sharp turns, and a yello line down the middle.

Place a series of circles along this chain, each having radius s, so that their centers are s units apart or less. In other words, the center of the "next" circle sits on the circumference of the previous circle.

Start at c, where f = g out to a distance of 2s. Move to the center of the next circle, which is s units away from c. Since f = g near this point, f = g out to a distance of 2s. Move to the center of the third circle, a distance of s from the second center. Once again f = g, extending out to a distance of 2s. This embraces the center of the fourth circle, and so on, all the way out to d. Once we reach d, f(d) = g(d).

If d is in the same connected domain as c, i.e. there is a path from c to d, the behavior of f at c completely determines the behavior of f at d.

If f is 0 near c, 0 spreads down the chain, and f is 0 everywhere.

In summary, an analytic function is completely defined, across its connected domain, by a sequence approaching a limit point c, or by its derivatives at c.

The only possible analytic extension of Ex in the complex plane is Ez. It is defined by the taylor series of Ex; just replace x with z.

If a relationship among analytic functions holds for real x, it extends to the entire domain. The function sin(z)2 + cos(z)2 -1 cannot be 0 on the x axis and nonzero elsewhere, so the familiar trig identity must be true over the complex plane.

If the condition holds then apply it to a real number x, and f conjugate of x = f(x), hence f(x) is real.

To show the converse, verify that f conjugate of z conjugate is a differentiable function iff f is differentiable at z. Write the difference quotient and take the limit as h approaches 0, and find the conjugate of f′, evaluated at the conjugate of z. I'll leave the algebra to you.

Let g = f on the upper half plane, and f conjugate of z conjugate on the lower half plane. Note that this definition is consistent on the x axis. Thanks to the above paragram, g is analytic above and below the x axis.

Let p be a real number, with g(p) = r. By assumption, r is also real. As we approach p from above, f approaches r, which means f conjugate also approaches r. Thus g approaches r from above and below, and g is continuous at p.

Since f is analytic, it is differentiable at p. As you recall, f′(p) times 1 gives the change in f as we move away from p in the x direction. Since f is real on the x axis, the change in f is real, and the derivative is real. Therefore g is differentiable from above, and g′(p) is a real number. The difference quotient from below approaches the same real value, so g is differentiable on the real line.

The extension of an analytic function is unique, so start with f on the upper half plane and extend to the lower half. This implies f = g = f conjugate of z conjugate.

For another proof, note that f′(x) is real for x real, and so on for the remaining derivatives. The power series consists of real coefficients. We may conjugate z, or the partial sum of the power series out to the nth term; the result is the same. Since partial sums are equal under pre and post conjugation, the same is true of f. Thus f conjugate of z = f of z conjugate.

To illustrate, let's apply this to Ez. The images Ea+bi and Ea-bi both lie on a circle, centered at the origin, with radius Ea. To find the first image, move around the circle in a counterclockwise direction, through an angle of b. To find the second image, move in a clockwise direction. The two images are symmetric about the real axis. In other words, they are conjugate.