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Expert: Steve Holleran - 5/6/2008


Question

Complex numbers Q14
Hi Steve, I've just been doing this complex numbers booklet and this question (attached) has stumped me. Please can you help me, my workings have got me no where.

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  

Steve Holleran

Expertise

I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience

33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

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