1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex Number Problem

  1. Sep 24, 2013 #1
    Two complex numbers z1 and z2 are taken such that |z1+z2|=|z1-z2|, and z2 not equal to zero.
    Prove that z1/z2 is purely imaginary (has no real parts).

    I started by taking z1=a+bi, and z2=c+di, then z1+z2=a+c+i(b+d) and z1-z2=a-c+i(b-d)

    |z1+z2|=√(a+c)^2 + (b+d)^2
    |z1-z2|=√(a-c)^2 + (b-d)^2

    As these don't equal each other, I need to choose another two complex numbers. But I'm not sure which ones to choose. I'm assuming this is the hardest part of the problem, and having found z1 and z2 I can divide them to show that no real parts remain. Any help is much appreciated, and thank you in advance.
    Last edited by a moderator: Sep 24, 2013
  2. jcsd
  3. Sep 24, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think you're confused - the point is that these typically aren't equal to each other, but if they accidentally happen to be, then you get some additional conditions on a, b, c and d that will let you prove that z1/z2 is purely imaginary.

    For starters, I would recommend squaring both sides. Then the condition |z1+z2| = |z1-z2| can be re-written as
    (a+c)2 + (b+d)2 = (a-c)2 + (b-d)2.

    Before you start doing algebra, it might help to figure out what the relationship between a, b, c and d are that will make z1/z2 purely imaginary.

    Also, I would recommend drawing a picture for this. In the complex plane, put a dot for z1. What are the (geometric) conditions for z2 that will make |z1+z2| = |z1 - z2|,, what are the geometric conditions for z1/z2 to be purely imaginary? This requires a reasonable amount of knowledge/familiarity with geometry in the complex plane so these might be difficult to answer.
  4. Sep 24, 2013 #3


    Staff: Mentor

    But what conditions on a, b, c, and d make it so that they are equal? You don't need two more complex numbers.
  5. Sep 24, 2013 #4


    User Avatar
    2017 Award

    Staff: Mentor

    There is a nice division which simplifies the problem significantly, and the geometric interpretation can help to find that.

    Hint: how would you prove it if z1 was known to be real?
  6. Sep 24, 2013 #5
    Hi Office_Shredder,

    If I make z1=a and z2=bi, then both the sum and the difference of both look like a complex number and its conjugate respectively. Also, |z1+z2|=|z1-z2| .

    z1/z2 then becomes a/bi. Multiplying top and bottom by bi produces -(a/b)i, which is only imaginary.

    Is this correct?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted