(4) Consider \(f(z) = \frac{(z^2 - 4)(z^2 - z - 12)}{(z^2 - 5z + 6)(z^2 - 9)^3}\) (a) What are the singularities of f? (b) Use Theorem 2 to determine which ones of singularities are poles?
Added by Nicholas M.
Close
Step 1
The denominator is $(z^2 - 5z + 6)(z^2 - 9)^3 = (z-2)(z-3)(z-3)^3(z+3)^3 = (z-2)(z-3)^4(z+3)^3$. The zeros are $z=2$, $z=3$, and $z=-3$. Therefore, the singularities are $z=2$, $z=3$, and $z=-3$. Show more…
Show all steps
Your feedback will help us improve your experience
Adi S and 54 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
3. For each of the following, construct a function f, analytic in the plane except for isolated singularities, that satisfies the given conditions. (a) f has a zero of order 2 at z = i and a pole of order 5 at z = 2 - 3i. (b) f has a simple zero at z = 0 and an essential singularity at z = 1. (c) f has a removable singularity at z = 0, a pole of order 6 at z = 1, and an essential singularity at z = i. (d) f has a pole of order 2 at z = 1 + i and essential singularities at z = 0 and z = 1.
Adi S.
For each of the following functions, say whether the indicated point is regular, an essential? singularity, or a pole, and if a pole of what order it is. (a) $\frac{\sin z-2}{z^{6}}, z=0$ (b) $\frac{z^{2}-1}{\left(z^{2}+1\right)^{2}}, z=i$ (c) $z e^{t / z}, \quad z=0$ (d) $\Gamma(z), \quad z=0)$ [See Chapter 11, equation (4.1).]
FUNCTIONS OF A COMPLEX VARIABLE
Laurent series
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD