What Are the Loci of Points Satisfying Complex Inequalities in the Plane?

saraaaahhhhhh
Messages
21
Reaction score
0

Homework Statement



#16)What are the loci of points z which satisfy the following relations...?

d.) 0 < Re(iz) < 1 ?

g.) α < arg(z) < β, γ < Re(z) < δ, where -π/2 < αα, β < π/2, γ > 0 ?

I'm also wondering for help with this proof:
#15)...Given:
z_1 + z_2 + z_3 = 0 and |z_1| + |z_2| + |z_3| = 1,
prove this defines an equilateral triangle inscribed in the unit circle |z| = 1. Any hints? What should I show to prove this?

Thanks!

(PS - the book is available on Google Books for those interested in seeing the original problems. They're on page 9, numbers 15/16. Book is by Silverman and called Introductory Complex Analysis.)


The Attempt at a Solution


For #15, I thought I might be able to use the inequality relations for triangles of a + b < c, or two sides are always less than the length of the third side, but that got me nowhere.
Then I thought I might be able to use arg(z_1) + arg(z_2) + arg(z_3) = pi, but I can't figure out what to do with that or how to prove it.

For #16, I just need clarification. Does 'loci of pts z which satisfy...' mean the shape that this set of points inscribes?

I just want to make sure I understand the problem. Even if I am right, though, I'm not sure how to approach these! I know that's not much of an attempt but even a hint or guideline would be helpful. z can be so many things...

Thanks again!

 
Physics news on Phys.org
Thanks for the help; I got the equilateral triangle one, but I'm still trying to figure out the other one.

How do I determine the loci of points for #16d and g?
 
saraaaahhhhhh said:
#15)...Given:
z_1 + z_2 + z_3 = 0 and |z_1| + |z_2| + |z_3| = 1,
prove this defines an equilateral triangle inscribed in the unit circle |z| = 1. Any hints? What should I show to prove this?
The question should read |z_1| = |z_2| = |z_3| = 1 !

saraaaahhhhhh said:
#16)What are the loci of points z which satisfy the following relations...?

d.) 0 < Re(iz) < 1 ?

g.) α < arg(z) < β, γ < Re(z) < δ, where -π/2 < αα, β < π/2, γ > 0 ?

The Attempt at a Solution


For #16, I just need clarification. Does 'loci of pts z which satisfy...' mean the shape that this set of points inscribes?
It means to describe (via a sketch and/or constraint inequations/equations of the components of z) the region in the complex plane in which points satisfy the given constraints. For example, for 16(d) we wish to describe the region in the plane in which points z satisfy 0<Re(iz)<1. Putting z = x + iy where x and y are real, you can show yourself this implies that 0 < -y < 1 (i.e., ...); and x is any real number.

Let's see your effort for (g). The first constraint can be handled by using the polar form of z and the latter using the rectangular form of z. (Note also that, for example, if alpha>-pi/2 then the first inequality additionally implies that pi/2>beta>-pi/2.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top