# Complex analysis: Find domain of definition and range

• Feb 16th 2009, 11:36 AM
srw899
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
• Feb 16th 2009, 12:07 PM
Jhevon
Quote:

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.

Quote:

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)

Quote:

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?

Quote:

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?
• Feb 16th 2009, 02:04 PM
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 \$\displaystyle z^2+ 1\$ 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 \$\displaystyle z^2+ 1\$ in the denominator.
• Feb 16th 2009, 02:13 PM
Jhevon
Quote:

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 \$\displaystyle z^2+ 1\$ 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 \$\displaystyle z^2+ 1\$ 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.
• Feb 18th 2009, 07:05 PM
ThePerfectHacker
Quote:

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)