Thread: Complex analysis: Find domain of definition and range

Find the domain of definition of each function:

a. f(z)=3z^2+5z+i+1

b. g(z)=1/z

c. h(z)=(z+i)/(z^2+1)

d. q(z)=(2z^2+3)/(|z-1|)

e. F(z)=e^3z

f. G(z)=e^z+e^-z

Describe the range of each function:

g. f(z)=z+5 for Re z>0

h. g(z)=z^2 for z in the first quadrant, Re z >=0, Im z >=0

i. h(z)= 1/z for 0<|z|<=1

j. p(z)=-2z^3 for z in the quarter-disk |z| < 1, 0<Arg z<pi/2

Originally Posted by srw899

Find the domain of definition of each function:

a. f(z)=3z^2+5z+i+1

b. g(z)=1/z

c. h(z)=(z+i)/(z^2+1)

d. q(z)=(2z^2+3)/(|z-1|)

e. F(z)=e^3z

f. G(z)=e^z+e^-z

the domains here will be pretty much the same as if z was a real number.

Describe the range of each function:

g. f(z)=z+5 for Re z>0

h. g(z)=z^2 for z in the first quadrant, Re z >=0, Im z >=0

Hint: when the domain is not restricted, polynomials are onto the whole complex plane (consequence of the Fundamental theorem of Algebra)

i. h(z)= 1/z for 0<|z|<=1

consider, regarding the range: what happens when z approaches zero? what happens if it is on the unit disk?

j. p(z)=-2z^3 for z in the quarter-disk |z| < 1, 0<Arg z<pi/2

see my comment for (g) and (h)


now, what can you come up with?

jhevon said "the domains here will be pretty much the same as if z was a real number."

Pretty much. One important difference is that is never 0 in the real numbers but is in the complex numbers it is. So in the complex numbers you have to watch out for fractions with in the denominator.

Originally Posted by HallsofIvy

jhevon said "the domains here will be pretty much the same as if z was a real number."

Pretty much. One important difference is that is never 0 in the real numbers but is in the complex numbers it is. So in the complex numbers you have to watch out for fractions with in the denominator.

indeed. i didn't notice part (c), i should have been more explicit. but i did mention the fundamental theorem of algebra, so we know that all polynomials have zeros, and dividing by zero is always a no-no. so we have to watch out if there is any polynomial in the denominator of a rational function.

Originally Posted by Jhevon

Hint: when the domain is not restricted, polynomials are onto the whole complex plane (consequence of the Fundamental theorem of Algebra)

Actually non-constant polynomials, but this trivial point was not why I wanted to responded. In fact any entire function (analytic everywhere) has range the whole complex plane (with an exception of only a single point)! This is such a deep, elegant and amazing result from complex analysis! (Picard's Little Theorem)