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Complex number homework

  1. Jul 18, 2014 #1
    1. The problem.

    Given that z= 3-4i
    Show that z^2 = 3-4i

    Hence or otherwise find the roots of the equation (z+i)^2=3-4i

    2. My attempt.

    The first part of the problem is strait forward z^2= (2-i)(2-i) then expand to get the desired result.
    Now the second part

    (z+i)^2=3-4i. Becomes

    z^2+ 2zi+i^2 = 3-4i

    From here on I replace z with 2-i and get nowhere!
     
  2. jcsd
  3. Jul 18, 2014 #2

    maajdl

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    Gold Member

    Given that z= 3-4i
    leads to z² = -7-24i
    There must be a mistake in your statement.

    Besides that, the equation z²=3-4i has two solutions:

    z1=(2-i) and z2=-(2-i)

    Solving (z+i)²=3-4i is then straightforward.
     
  4. Jul 18, 2014 #3
    Yes there was a mistake! It should have read... Given that z = 2-i
     
  5. Jul 18, 2014 #4
    So you have (2 - i)2 = 3 - 4i and you are looking for roots for (x + i)2 = 3 - 4i (I've changed the unknown to x to avoid confusion).
     
  6. Jul 18, 2014 #5
    Yes! The variable z does confuse things a bit.
     
  7. Jul 18, 2014 #6

    SammyS

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    Staff Emeritus
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    Don't forget, if you use x this way it is possibly a complex number !
     
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