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B Complex number solutions

  1. Feb 28, 2017 #1
    This is a question from a competitive entrance exam ....I just want to check whether my approach is correct as i dont have the answer keys .

    here is the question :

    How many complex numbers z are there such that |z+ 1| = |z+i| and |z| = 5?
    (A) 0
    (B) 1
    (C) 2
    (D) 3

    My approach :
    let z = x+iy
    Now, using |z+ 1| = |z+i|,
    |(x+1)+iy| = |x+(y+1)i|
    Simplifying this, i got x=y......(1)

    and since |z| = 5 , we have √(x2+y2) = 5
    which means (x2+y2) = 25 ......(2)

    Now, plugging (1) in (2) , we get

    x2 = (25/2)

    therefore x can take 2 values similarly y also can take 2 values.....
    and since x=y in the complex number .....we have 2 solutions and hence the answer is 2
  2. jcsd
  3. Feb 28, 2017 #2


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    Hello m1, :welcome:

    I would do it exactly the same way. Does that help ?
    (An alternative, graphical approach is to draw a circle in the complex plane and pick the two points that have this property)
  4. Feb 28, 2017 #3
    Thank you sir .....but just a thought about your alternative ,
    i would have to pick points in the circle centered at origin having radius 5 which are equidistant from (0,1) and (1,0) ....but doesnt that persues the same algebraic calculations ? So its not an easier alternative in any sense right ?
  5. Feb 28, 2017 #4


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    for which a translation +1 on the real axis ends up at the same distance from the origin as a translation +1 on the imaginary axis
    And yes, it's the same thing as what you do algebraically.
  6. Mar 28, 2017 #5


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    Yes, except that the possible z solutions should be equidistant from -1 and -i. It's lucky that your answers are the same.
  7. Mar 28, 2017 #6


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    Following up on FactChecker's post, the number lies on the bisector, i.e., the line bisecting the segment from -1 to i , which is the x-axis. You can see that there are two solutions.
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