Complex Analysis Assignment Solutions: Unit Circle, Continuity Proofs

This document presents detailed solutions to a complex analysis assignment, covering a range of topics including finding values for complex functions, evaluating contour integrals using parameterization and the residue theorem, proving the continuity and analyticity of functions like sin(z), and exploring properties of functions such as z = tan(w). The solutions demonstrate techniques for evaluating integrals along different paths, applying the Cauchy-Riemann equations, and using Liouville's theorem to determine the nature of complex functions. The assignment delves into concepts like poles, residues, and domains of continuity, providing a comprehensive overview of essential complex analysis principles. Desklib offers a platform to access similar solved assignments and study resources for students.

Solution 1: We need to find all values of c for which
.
Since,
.
Now
Therefore is not possible.
Now let , then
Let , then
Let , then
Similarly for
Hence, for , where C is the set of complex number of the form , a is
any real number and .
Solution 2:
2(a): .Consider C1. Along AB, and x varies from -1 to 1.
Since and . So
Along BAC: and y varies from 0 to 1. Let . Since
, where and .
So,
Hence,
2(b): .Consider C2. Along PR, and y varies from 0
to 1.
Since and . So

Along RS: and x varies from -1 to 1.
Since .
So,
Along SQ: and y varies from 1 to 0.
Since and . So
Hence,
2(c): .Consider C1. Along AB, and x varies from -1 to 1.
Since and . So
Along BAC: and y varies from 0 to 1. Let . Since
, where and .
So,
Hence,
2(d): .Consider C2. Along PR, and y varies from 0
to 1.
Since and . So
Along RS: and x varies from -1 to 1.
Since .
So,
Along SQ: and y varies from 1 to 0.
Since and . So
Hence,

Solution 3: Given that C is the unit circle that is . We need to find , where
. Since, , suppose , where .
Now
So,
Hence,
How for ?
We need to find
Since pole of the function is z = 0
And z = 0 lies on the unit circle . And residue at z = 0 is 1. So by residue theorem
Solution 4: To prove sin z continuous on C.
Suppose .|, that is z is inside a disc of radius and center . By geometry we
have
Now,
That is for any , chose , then
. Hence by definition sin z is continuous on C.
To prove sin z is analytic on C.
Since, . We know that sum of two analytic function is analytic. So to
prove sin z is analytic on C, it is suffices to show that eiz and e-iz are analytic. Note that a
function f (z) = u + iv is analytic if it satisfy C – R equation. That is
and .
Let
This implies that . Now differentiate u with respect to x and y
partially, we get,

…… (1)
Differentiate v with respect to x and y partially, we get,
….. (2)
From equation (1) and (2) we get,
and .
Similarly we can prove that is also analytic. Hence is analytic. This
complete the proof.
To find derivative of sin z .
Since, so,
Solution 5(a): To prove . Given z ∈ C, we need to find w ∈ C such
that z = tan (w). That is,
Now, let’s find the value of w. The above equation can be written as
This implies that,
Since z = tan w this implies . Hence from last equation of above equation we
get,
This completes the proof.
5(b) Since and is continuous over C except . So the
largest domain in which continuous is
5(c): Since