Analytic Functions : Constancy
7. Let a function f (z) be a analytic in a domain D. Prove that f (z) must be constant throughout D if (a) f (z) is real-valued for all z in D (b) | f (z) | is constant throughout D. (Question also included in attachment)
Show that u (x, y) is harmonic in some domain and find a harmonic conjugate v (x, y)
1. Show that u (x, y) is harmonic in some domain and find a harmonic conjugate v (x, y) when (a) u (x, y) = 2x (1 – y) (b) u (x, y) = 2x – x3 + 3xy2 (c) u (x, y) = sinh x•sin y (d) u (x, y) = y / (x2 + y2) (Question is also included in attachment)
Complex Variables : Harmonic Conjugates
Show that if v and V are harmonic conjugates of u in a domain D, then v (x, y) and V (x, y) can differ at most by an additive constant.
Cauchy-Riemann Equations : First-Order Partial Derivatives
Please see the attached file for full problem description. 3. Use Cauchy-Riemann equations and the given theorem to show that the function _ f (z) = e^z is not analytic anywhere. Theorem: Suppose that f (z) = u (x, y) + i v (x, y) and that f'(z) exists at a point z0 = x0 + i y0. Then the first- ...continues
Please see the attached file for full problem description. --- 5. Write |exp (2z + i) | and | exp (iz2) | in terms of x and y. Then show that | exp (2z + i) + exp (iz2) | ≤ e2x + e-2xy. (exp means exponential function)
Please see the attached file for full problem description. --- 3. Show that (a) Log (1 + i)2 = 2 Log (1 + i) (b) Log (-1 + i)2 ≠ 2 Log (-1 + i).
Verify log(z1/z2) = log z1 - log z2 by a) using the fact that arg(z1/z2) = arg z1 - arg z2 b) showing that log(1/z) = -log z... Please see the attached file for the fully formatted problems.
Show that z^(1/n) = exp((1/n)log z) also holds when n is a negative integer.... Please see the attached file for the fully formatted problems.
Find the principal value of... Please see the attached file for the fully formatted problems.
Prove that if all of the powers invoved are principal values then... Please see the attached file for the fully formatted problems.