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Powers of a Complex Number Problem

  • Thread starter Bng1290
  • Start date
  • #1
5
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Homework Statement



I'm pretty sure that I just don't fully understand these problems so I think I just need help getting pushed in the right direction here. Anyways, here's the problem I'm on.

Evaluate and give your answer in Cartesian Coordinates:
[tex]\left|(1-3i)^5(\sqrt{2}+i\sqrt{3})^7\right|[/tex]

Homework Equations



[tex]\left|z\right|[/tex]= r = sqrt(x^2+y^2)
z=x+iy=re^(i[tex]\Theta[/tex])
z^n=r^n e^(in[tex]\Theta[/tex])

The Attempt at a Solution



I've tried to convert to polar form (re^i[tex]\Theta[/tex]) and raising r to the nth power and multiplying the angle by n for each of the two separate complex numbers and then multiplying together from there. Then I assume I would have to convert back to cartesian coords for the answer. I honestly think I'm missing something important with all of this as I thought that the absolute value of a complex number was its modulus and thus I would not be able to get any answer that could be presented in cartesian coordinates. I've read through my book on this subject and can't really figure out what to do, which is what I meant when I said I need help getting in the right direction.
 

Answers and Replies

  • #2
Dick
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Since the angles don't come out to be nice angles, I think you are better off staying in cartesian coordinates. Just multiply it out.
 
  • #3
fzero
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Just to add to Dick's suggestion. In case it's not already clear to you, the modulus of a complex number satisfies

[tex]|z^p| = |z|^p,[/tex]
[tex]|wz| = |w||z|,[/tex]

which can be used to break your calculation up into steps until you're clear how it all works.
 
  • #4
5
0
Okay so I went ahead and multiplied everything out and I got

[tex]\left|(316+12i)(197\sqrt{2}-13\sqrt{3}i)\right|[/tex]

I still don't really know what to do from here, considering this is all supposed to be doable without a calculator I have a feeling this must not be the right way with all the heinous math involved (especially if I multiply these two together in the next step). And ultimately I still don't know how my answer will be in x+iy form if this is an absolute value. I think I'm just terribly confused about all of this.
 
  • #5
Dick
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Now I'm a little confused. They asked you to give the answer in cartesian coordinates, so I assumed they just wanted the product. It's a lot easier to just give the absolute value as fzero pointed out. So I'm not that sure what they actually want either. But the next step in multiplying it out shouldn't be that bad. But how did you get all those funny numbers in the second factor?
 
  • #6
12
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Okay so I went ahead and multiplied everything out and I got

[tex]\left|(316+12i)(197\sqrt{2}-13\sqrt{3}i)\right|[/tex]

I still don't really know what to do from here, considering this is all supposed to be doable without a calculator I have a feeling this must not be the right way with all the heinous math involved (especially if I multiply these two together in the next step). And ultimately I still don't know how my answer will be in x+iy form if this is an absolute value. I think I'm just terribly confused about all of this.
I just checked the above expression and you did multiply correctly. It is much easier to just use the properties of the modulus instead though.

[tex]\left|(1-3i)^5(\sqrt{2}+i\sqrt{3})^7|= |1-3i|^5|\sqrt{2}+\sqrt{3}i\right|^7
\left=(\sqrt{1^2+(-3)^2} )^5(\sqrt{\sqrt{2}^2+\sqrt{3}^2 )^7\right
\left=(\sqrt{10})^5(\sqrt{5})^7=62500\sqrt{2}\right[/tex]

This is all easily done without the use of a calculator. Your way involved many more tedious calculations. I would neglect the cartesian coordinate comment. The modulus is not a coordinate in the cartesian plane but a measure of how far any complex number is from the origin.

Sorry my exponets look a little messed up. It should be the whole squareroot raised to the 5th and 7th power.
 
  • #7
5
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So I've decided to just ignore the the comment about cartesian coordinates and to be satisfied with solving for the modulus, which I was able to do. Thanks for the help everyone.
 
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