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Complex Numbers- Square root 3i

  1. Aug 12, 2009 #1
    (1- sqrt 3i) ^3

    I am having trouble solving the sqrt 3i part. I think I need to use de moivres theorem but I am unsure. If someone could push me in the right direction that would be a massive help. Thanks.
  2. jcsd
  3. Aug 12, 2009 #2


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    What's the question exactly? You want to write that complex number in its polar representation or as x+iy? Either way start with writing [itex]1-\sqrt{(3i)}[/itex] in the polar representation. Once you have that it is easy to compute [itex](1-\sqrt{(3i)})^3[/itex].
    Last edited: Aug 12, 2009
  4. Aug 12, 2009 #3
    I think he meant:


    Use De Moivre's theorem to find all the solutions.
  5. Aug 12, 2009 #4
    The question is:

    If z= 5 +4i, write the number z(|z|^2-(1- sqrt3i)^3) in the form a+bi.

    I am able to sub everything in but I want to simplify the (1- sqrt3i)^3. The square root has thrown me off.
  6. Aug 12, 2009 #5


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    I take it the i is inside the square root? Try to write it in the form [itex]3i=|z|e^{i\phi}[/itex] then take the square root on both sides.
  7. Aug 12, 2009 #6
    Is that meant to the theta or phi? Silly question I know. I have in my notes a simalar formula for Eulers formula but that uses theta not phi.... Another silly question I know but by changing the sq root 3i into a simple number going to give me the same result as getting the polar representation of the whole (1 - sqrt3i) ^3?
  8. Aug 12, 2009 #7


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    It doesn't matter whether it's called phi, theta or JC_003 it is just a variable which represents the angle between |z| and the positive real axis. You will get the same answer both ways, as it should. However the angle theta is pretty hard to find for [itex]1-\sqrt{3i}[/itex]. I suggest you write [itex]\sqrt{3i}=x+iy[/itex] first and then continue from there.
  9. Aug 12, 2009 #8
    If this is an exercise from a text, I'd be very carefull about whether the question concerns




    Huge difference.

    The question is not clear enough for me to help effectively, but I suspect -8 shows up somewhere.

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