Let f be an entire function such that |f(z)|<=A|z|. Use Cauchy's inequality to show that f(z)=az for some complex constant a.

Let f be an entire function such that |f(z)|<=A|z|. Use Cauchy's inequality to show that f(z)=az for some complex constant a. See the attachment for a more complete description of the question and Cauchy's inequality.

Show that the minimum of a non-constant, non-zero analytic function on a closed bounded region occurs strictly on the boundary of the region.

See attached PDF for a full version of the question with correct mathematical notation.

Maximum Modulus Principle

See attached document. Let f be the function f(z)=e^z and R the rectangular region....

An example of contour integration using Cauchy's formula

Evaluate the contour integral of z^2/(4-z^2) around the circle |z+1|=2. The question is attached in correct mathematical notation, along with the student's (incorrect) initial attempt. You will need to refer to this initial attempt when reading the solution.

Find the max and min modulus if f(x) = z + 3i on the closed region defined by |z| < 2.

Find the max and min modulus if f(x) = z + 3i on the closed region defined by |z| < 2. I don't have clue as to where to start.

deMoivre's Theorem

Please see the attached file.

Find the equation of parabola

#1. Find the equation of parabola describe. Find 2 points of latus rectum.Graph. Focus(-5,0) Vertex(0,0) #2 Find the equation of the parabola. Find 2 points that define latus rectum. Graph. Focus (0,1) Diectrix line y= -1 #3. Find the equation of ellipse.draw the graph. Center (0,0) Focus(0,8) Vertex ( ...continues

Suppose f and g are functions analytic in a domain D.

Suppose f and g are functions analytic in a domain D. If z_n is a bounded sequence of distinct points in D and if f(z_n) = g(z_n) for all n, show that f(z) = g(z) for all z in D. Is the same true for an unbounded sequence?

Analytic

Suppose that f is analytic in the disc |z|<1, that f(0) = f’(0) = 0 and that |f(z)| ≤ 1 for all z in the disc. Show that |f(z)| ≤ |z^2|, for |z|<1. Hint: Schwarz’s Lemma: Suppose that f is analytic in the disc |z|<1, that f(0) = 0 and that |f(z)| ≤ 1 for all z in the disc. Then |f(z)| ≤ |z|, for |z| ...continues

Complex analysis conformal mapping

a) find a bijective conformal mapping that takes a bounded region to an unbounded region b) prove that a conformal map cannot take a simply connected region onto a region that is not simply connected.