Complex Differentiation Mohammed Nasser Department of statistics RU

Derivatives Differentiation of complex-valued functions is completely analogous to the real case: Definition. Derivative. Let f(z) be a complex-valued function defined in a neighborhood of z 0. Then the derivative of f(z) at z 0 is given by Provided this limit exists. f(z) is said to be differentiable at z 0.

Some Exercises Show that 1)f(z)= is nowwhere differentiable. 2)g(z)=z n has derivative nz n-1. 3)h(z)=e z has derivative e z. 4)l(z)=|z| 2 is nowhere differentiable except z=0 5)Every real-valued function of complex variable is either non-differentiable or differentiable with derivative equal to 0.

Solutions 1. h 0 X- axis Y- axis If we go along X-axis A tends to 1. A If we go along Y-axis A tends to -1. That implies the limit does not exist.

Properties of Derivatives

Analytic. Holomorphic. Definition. A complex-valued function f (z) is said to be analytic, or equivalently, holomorphic, on an open set  if it has a derivative at every point of . (The term “regular” is also used.) It is important that a function may be differentiable at a single point only. Analyticity implies differentiability within a neighborhood of the point. This permits expansion of the function by a Taylor series about the point. If f (z) is analytic on the whole complex plane, then it is said to be an entire function.

Rational Function. Definition. If f and g are polynomials in z, then h (z) = f (z)/g(z), g(z)  0 is called a rational function. Remarks. –All polynomial functions of z are entire. –A rational function of z is analytic at every point for which its denominator is nonzero. –If a function can be reduced to a polynomial function which does not involve, then it is analytic.

Example 1 Thus f 1 (z) is analytic at all points except z=1.

Example 2 Thus f 2 (z) is nowhere analytic.

Testing for Analyticity Determining the analyticity of a function by searching for in its expression that cannot be removed is at best awkward. Observe: It would be difficult and time consuming to try to reduce this expression to a form in which you could be sure that the could not be removed. The method cannot be used when anything but algebraic functions are used.

Cauchy-Riemann Equations (1) If the function f (z) = u(x,y) + iv(x,y) is differentiable at z 0 = x 0 + iy 0, then the limit can be evaluated by allowing  z to approach zero from any direction in the complex plane.

Cauchy-Riemann Equations (2) If it approaches along the x-axis, then  z =  x, and we obtain But the limits of the bracketed expression are just the first partial derivatives of u and v with respect to x, so that:

Cauchy-Riemann Equations (3) If it approaches along the y-axis, then  z = i  y, and we obtain And, therefore

Cauchy-Riemann Equations (4) By definition, a limit exists only if it is unique. Therefore, these two expressions must be equivalent. Equating real and imaginary parts, we have that must hold at z 0 = x 0 + iy 0. These equations are called the Cauchy-Riemann Equations. Their importance is made clear in the following theorem.

Cauchy-Riemann Equations (5) Theorem. Let f (z) = u(x,y) + iv(x,y) be defined in some open set  containing the point z 0. If the first partial derivatives of u and v exist in , and are continuous at z 0, and satisfy the Cauchy-Riemann equations at z 0, then f (z) is differentiable at z 0. Consequently, if the first partial derivatives are continuous and satisfy the Cauchy-Riemann equations at all points of , then f (z) is analytic in .

Example 1 Hence, the Cauchy-Riemann equations are satisfied only on the line x = y, and therefore in no open disk. Thus, by the theorem, f (z) is nowhere analytic.

Example 2 Prove that f (z) is entire and find its derivative. The first partials are continuous and satisfy the Cauchy-Riemann equations at every point.

Harmonic Functions Definition. Harmonic. A real-valued function  (x,y) is said to be harmonic in a domain D if all of its second-order partial derivatives are continuous in D and if each point of D satisfies Theorem. If f (z) = u(x,y) + iv(x,y) is analytic in a domain D, then each of the functions u(x,y) and v(x,y) is harmonic in D.

Harmonic Conjugate Given a function u(x,y) harmonic in, say, an open disk, then we can find another harmonic function v(x,y) so that u + iv is an analytic function of z in the disk. Such a function v is called a harmonic conjugate of u.

Example Construct an analytic function whose real part is: Solution: First verify that this function is harmonic.

Example, Continued Integrate (1) with respect to y:

Example, Continued Now take the derivative of v(x,y) with respect to x: According to equation (2), this equals 6xy – 1. Thus,

Example, Continued The desired analytic function f (z) = u + iv is:

Remember Complex Exponential We would like the complex exponential to be a natural extension of the real case, with f (z) = e z entire. We begin by examining e z = e x+iy = e x e iy. e iy = cos y + i sin y by Euler’s and DeMoivre’s relations. Definition. Complex Exponential Function. If z = x + iy, then e z = e x (cos y + i sin y). That is, |e z |= e x and arg e z = y.

More on Exponentials Recall that a function f is one-to-one on a set S if the equation f (z 1 ) = f (z 2 ), where z 1, z 2  S, implies that z 1 = z 2. The complex exponential function is not one-to-one on the whole plane. Theorem. A necessary and sufficient condition that e z = 1 is that z = 2k  i, where k is an integer. Also, a necessary and sufficient condition that is that z 1 = z 2 + 2k  i, where k is an integer. Thus e z is a periodic function.

How is the case with multi-valued functions like z 1/n, logz etc??