Answer
Hi Jaime,

Well, I think I have something put together here.  Let's take a look:

Since z = a + bi, then z + i = a + bi + i and

|z + i| = |a + (b+1)i| = sqrt(a^2 + (b+1)^2) = 3

and then                a^2 + (b+1)^2 = 9

so                      a^2 + b^2 + 2b + 1 = 9

and                     a^2 + b^2 + 2b = 8.     ***



Also, from above arg z-1 = arg (a+bi-1) = arg(a-1 + bi)

     arg(z-1) = arctan(b/(a-1)) = pi/4

means that         b/(a-1) = tan pi/4 = 1

so b = a-1.***


So to solve for the intersection of the two sets, we solve the *** equations simultaneously.  I think substitution is pretty easy here:

 a^2 + (a-1)^2 + 2(a-1) = 8

 a^2 + a^2 -2a + 1 + 2a - 2 = 8

 2a^2 -1 = 8    so 2a^2 = 9  , a^2 = 9/2  so a =+/- 3/rt(2) or

 a = +/- 3*rt(2)/2.  Then b = a - 1 = (+/-3rt(2) - 2) /2.

Then just put these expressions in the a + bi form:

z = +/- 3rt(2)/2 +/- (3rt(2)-2)/2 * i


I hope this is what you needed.

Steve  

Add to this Answer    Ask a Question

Rate this Answer

Was this answer helpful?

Not at all Definitely                  1 2 3 4 5

Related Articles • Biographies of Notable Women of the Ancient and Classical World • Biographies of Notable Medieval Women • Die Hard: Vendetta Cheat Codes - Gamecube • Alternative Therapy Directory -- A-Z Guide to Alternative Therapies on About.com • Women's History Picture Gallery - Individuals