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Ablowitz & Fokas Secs. 1.3,2.1 Ahlfors Secs. 2.1.1, 2.1.2 Dettman Secs. 2.2, 2.3, 2.5 Reading: Homework 1 due Friday, September 27 at 5 PM. It suffices to show the proposition for the special case in which the reference circle C is the unit circle. (If it's true for the unit circle, then simply by dilation and translation the unit circle can get mapped to any other reference circle, and the properties of the proposition are clearly invariant under dilation and translation.) Under stereographic projection, the unit circle maps to the equation of the Riemann sphere. Two symmetric points on the complex plane (relative to the unit circle) map to two points that are north-south symmetric on the Riemann sphere. Two circles on the Riemann sphere intersect in points that are north-south symmetric (symmetric w.r.t. the equatorial plane) if and only if both circles are orthogonal to the equator. Now doing reverse stereographic projection onto the complex plane, we obtain the statement of the proposition because stereographic projection is isogonal and circular. We can prove the proposition stated at the end of the previous lecture rather elegantly using stereographic projection. The proposition above can be shown to be true as well if the points are symmetric with respect to a line rather than a circle. (Elementary geometry proves it.) Any line (not a circle) in the complex plane that passes through two points that are symmetric w.r.t. the unit circle corresponds to a great circle on the Riemann sphere under stereographic projection. A few comments: The key useful proposition that follows from consideration of reflections across and lines and circles is: Proposition: Two points that are symmetric with respect to a line or circle will be mapped, under any LFT, to two points that are symmetric w.r.t the image of the line or circle. Proof: LFTs are isogonal, in fact conformal, so properties of angles of Complex Differentiability Monday, September 23, 2013 1:58 PM ComplexAnalysis Page 1

Ablowitz & Fokas Secs. 1.3,2.1 Homework 1 due Friday ...homepages.rpi.edu/~kramep/ComplexAnalysis/Notes/canotes092313.pdf · •Ablowitz & Fokas Secs. 1.3,2.1 ... Homework 1 due Friday,

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Page 1: Ablowitz & Fokas Secs. 1.3,2.1 Homework 1 due Friday ...homepages.rpi.edu/~kramep/ComplexAnalysis/Notes/canotes092313.pdf · •Ablowitz & Fokas Secs. 1.3,2.1 ... Homework 1 due Friday,

Ablowitz & Fokas Secs. 1.3,2.1•Ahlfors Secs. 2.1.1, 2.1.2•Dettman Secs. 2.2, 2.3, 2.5•

Reading:

Homework 1 due Friday, September 27 at 5 PM.

It suffices to show the proposition for the special case in which the reference circle C is the unit circle. (If it's true for the unit circle, then simply by dilation and translation the unit circle can get mapped to any other reference circle, and the properties of the proposition are clearly invariant under dilation and translation.)

Under stereographic projection, the unit circle maps to the equation of the Riemann sphere. Two symmetric points on the complex plane (relative to the unit circle) map to two points that are north-south symmetric on the Riemann sphere.

Two circles on the Riemann sphere intersect in points that are north-south symmetric (symmetric w.r.t. the equatorial plane) if and only if both circles are orthogonal to the equator.

Now doing reverse stereographic projection onto the complex plane, we obtain the statement of the proposition because stereographic projection is isogonal and circular.

We can prove the proposition stated at the end of the previous lecture rather elegantly using stereographic projection.

The proposition above can be shown to be true as well if the points are symmetric with respect to a line rather than a circle. (Elementary geometry proves it.)

Any line (not a circle) in the complex plane that passes through two points that are symmetric w.r.t. the unit circle corresponds to a great circle on the Riemann sphere under stereographic projection.

A few comments:

The key useful proposition that follows from consideration of reflections across and lines and circles is:Proposition: Two points that are symmetric with respect to a line or circle will be mapped, under any LFT, to two points that are symmetric w.r.t the imageof the line or circle.

Proof: LFTs are isogonal, in fact conformal, so properties of angles of

Complex DifferentiabilityMonday, September 23, 20131:58 PM

ComplexAnalysis Page 1

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Proof: LFTs are isogonal, in fact conformal, so properties of angles of intersection are preserved by the LFT. The previous proposition characterized symmetric points entirely in terms of properties of angles of intersection of lines and circles.

Why is this useful? If one knows what the image of a point should be and the image of a line or circle should be, then this proposition gives one more piece of specific information, namely where the symmetric points maps to.

Complex differentiability

One says that a function f(z) of a single complex variable is differentiable at a point z0 provided that:

exists and f'(z0)

The notion of differentiability of a complex variable, in this sense, is much stronger a restriction than differentiability for a function of two real variables; it requires more than smoothness.

A function is said to be analytic at z0 if it is differentiable in an open neighborhood of z0.

Appropriate way to define differentiability at is to say: f(z) is differentiable at if and only if f(1/z) is differentiable at 0 (actually to be more precise, if f(1/z) is differentiable or has a removable singularity at 0).

Not allowing the derivative to be infinite is actually important for the function to have nice properties. In fact, in some ways having a zero derivative is geometrically as bad as having an infinite derivative. But when one is doing calculus involving complex functions, zero derivative points don't affect the calculation but points with infinite derivatives do.

Comments about exclusion of

Let's explore why complex differentiability is stronger than simply assuming the function is smooth. The key observation is that in the limit, h is also a complex number, and so the existence of the limit actually means that the various directional derivatives of the function f(z) must be related to each other.

ComplexAnalysis Page 2

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f(z) = u(x,y) + i v(x,y) ○

Another way to see this is that if one speaks of a two-dimensional real vector field on the two-dimensional real plane, then differentiability of the function implies that it can be locally expressed to leading order in terms of a matrix of 4 partial derivatives that are typically independent of each other. But here, the complex derivative must be a complex number, which just has 2 real components, so the notion of complex differentiability must be more constrained than the definition of differentiability of multivariate real functions.

•derivatives of the function f(z) must be related to each other.

Let's examine what the extra constraints are by considering limits taken in

particular directions: h= (polar coordinates).

To understand what this implies in terms of the real and imaginary part of the complex variables let's write:

z0=x0+iy0

f(z) = u(x,y) + i v(x,y)

Notice that the particular limit we are taking does correspond to h :

Now let's assume that u,v are differentiable as real-valued functions, and multiply this expression by:

ComplexAnalysis Page 3

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expression by:

Where

But complex differentiability insists that the result of this limit, which

corresponds to sending h = should only depend on z0 and not the way in which the limit was taken. In particular, the result must be independent of , the direction in which z0 was approached in the limiting process.

This forces the partial derivatives of the real and imaginary parts to satisfy the Cauchy-Riemann conditions:

ComplexAnalysis Page 4

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because are linearly independent functions.

This leads us to a key proposition:A complex function f(z) is analytic at a point z0 if and only if its real and imaginary parts have continuous partial derivatives and satisfy the Cauchy-Riemann conditions at z0.

Proof: Necessity of Cauchy-Riemann conditions was shown above (modulo the technicality that the partial derivatives be continuous -- see the texts).

Sufficiency of Cauchy-Riemann conditions and continuous partial derivatives?This implies that:

ComplexAnalysis Page 5

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Further consequences:If f(z) is twice continuously differentiable and analytic, then the real and imaginary parts are harmonic as functions of two real variables.

This is just a corollary of the CR conditions:

We say that two functions u,v which satisfy the Cauchy-Reimann conditions and are twice continuously differentiable are conjugate harmonic.

So we can summarize the above by saying that two real functions of two real variables can serve as the real and imaginary parts of a complex analytic function if they are conjugate harmonic.

ComplexAnalysis Page 6