 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Region Area Ascertainment (Posted on 2014-12-28) 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.) My region isn't reaching (spoiler) | Comment 1 of 4
I think the area of R is 0.

Let z = a + bi.

In order for z/40 to have both parts between 0 and 1, a and b must both be between 0 and 40.

But 40/(a+bi) = 40*(a-bi)/((a+bi)*(a-bi) = 40(a-bi)/(a^2 + b^2).
Clearly, b must be non-positive in order for 40/z to have both parts between 0 and 1.  If b = 0, then 40/(a+bi) = 40/a

Putting these results together, a is in the range (0,40] and b = 0.  So R is a line segment with length = 40 and Area = 0.

Edited on December 28, 2014, 10:50 am
 Posted by Steve Herman on 2014-12-28 10:13:27 Please log in:

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