z solutions

All z solutions to z^3=9-13i

Complex Logarithm

Given: z1 = i and z2 = -1+i Show that: Log (z1 z2) ≠ Log z1 + Log z2

Evaluating Integral Using Jordan's Lemma

The problem is to find the value of the integral from 0 to INF of [(ln x)^2]/(x^2 +9). We are to use f(z)= [(Log z)^2]/(z^2 +9), where -pi/2 < Log z < 3pi/2. We are to use the curve C from -R to -p along the real axis, -p to p around 0, p to R along the real axis, and the curve Cr from 0 to pi. I am having several problems with this one, including: **finding the residue at 3i. **figuring out how to set up and equate the four integrals to 2pi(i) times the residue at 3i. Your help will be much appreciated, and if it is good help I have two more that I cannot do for which I will also gladly pay for some assistance. Our professor is retired and has no office hours, so I do n... click for more

Evaluating an Integral

The problem is: Evaluate the integral from 0 to INF of: [(x^(1/3))*(ln x)]/(x^2 +9) dx by using f(z)= [(z^(1/3))*(Log z)]/(z^2 +9), with -pi/2 < Log z < 3pi/2. Also, with z^(1/3)= e^[(1/3)Log z]. We are to use the curve C: from -R to -p, -p to p around origin, p to R, and Cr from 0 to pi. Many thanks in advance for your help. This is TOUGH one!

Evaluating an Integral With 2nd Order Pole

Problem: Evaluate the integral from 0 to INF of: (x^a)/(x^2 +4)^2 dx, -1 < a < 3 We are to use f(z)= (z^a)/(z^2 +4)^2, with z^a = e^(a Log z), Log z= ln|z| + i Arg z, and -pi/2 < Arg z < 3pi/2. I have found the residue at 2i to be: [2^a(1-a)/16]*[cos ((pi*a)/2) + i sin ((pi*a)/2). Please let me know if this is correct and how to solve this problem. Many thanks for your help.