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Complex Numbers- Square root 3i

  • Thread starter JC_003
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  • #1
3
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(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.
 

Answers and Replies

  • #2
Cyosis
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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:
  • #3
365
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I think he meant:

[tex]\sqrt[3]{1-i\sqrt{3}}[/tex]

Use De Moivre's theorem to find all the solutions.
 
  • #4
3
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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.
 
  • #5
Cyosis
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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.
 
  • #6
3
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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?
 
  • #7
Cyosis
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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.
 
  • #8
286
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If this is an exercise from a text, I'd be very carefull about whether the question concerns

[tex](\sqrt{3})i[/tex]​

or

[tex]\sqrt{(3i)}[/tex]​

Huge difference.

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

--Elucidus
 

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