All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Region Area Ascertainment (Posted on 2014-12-28) Difficulty: 3 of 5
R is the region in the complex plane that consists of all points z such that each of z/40 and 40/z have real and imaginary parts between 0 and 1 inclusively.

Find the area of R.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Another Solution Comment 4 of 4 |
Express z/40 in polar form, call it (r,theta).  For z/40 to be in the square, 0<r<=sqrt(2) and 0<=theta<=pi/2. (These bounnds are necessary but not sufficient.)

40/z is the reciprocal of z/40.  Then 40/z in polar form equals (1/r, -theta).  For 40/z to also be in the square, 0<1/r<=sqrt(2) and 0<=-theta<=pi/2.

The two theta ranges intersect at one value: theta=0.  That reduces the possible values of z/40 and 40/z to the real axis.  With this restriction then 0<{40/z and z/40}<1.  A positive number and its reciprocal can only be in that range if the number is 1.  Then 40/z=z/40=1.

Therefore the set of all z is the single real number z=40.

  Posted by Brian Smith on 2016-12-05 12:01:28
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information