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Finding polar form of 1/Z

  1. May 20, 2008 #1
    Hi There.

    I was given this question and the answer:

    Find the polar forms of [tex]1/z[/tex] where z = [tex]\sqrt{}3[/tex] + i


    and [tex]1/z[/tex] where z = 4[tex]\sqrt{}3[/tex] -4i

    Answers respectively are:

    [tex]1/2[/tex] cis([tex]-\pi/6[/tex])

    [tex]1/8[/tex] cis([tex]\pi/6[/tex])


    Can someone please explain to me why it is that the sign of the argument changes in the answer from + to -, for the first question, and - to +, for the second question?
     
  2. jcsd
  3. May 20, 2008 #2

    Dick

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    Because the argument of a+bi and the argument of a-bi are negatives of each other? Why? Of course they don't have to be. You could equally well have written the first one as (1/2)*cis(11*pi/6). Why?
     
    Last edited: May 20, 2008
  4. May 20, 2008 #3
    I understand that that is how the question is written and the sign indicates the point position in the plane. Sorry I worded the question wrong.

    Basically, z = [tex]\sqrt{}3[/tex] + i

    Polar form of this is: 2( cis [tex]\pi[/tex]/6 )

    then working out 1/z the answer is 1/2 (cis -[tex]\pi[/tex]/6)

    I do not know why the sign changes to make [tex]\pi[/tex]/6 negative and whether you take 1 as r, and z as t, and use the formula for division which is r/t ( cos ([tex]\pi[/tex] - [tex]\phi[/tex]) + isin ([tex]\pi[/tex] - [tex]\phi[/tex]) but when I tried that I still could not come up with a logical explanation it seemed.
     
  5. May 20, 2008 #4

    Defennder

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    [tex]\frac{1}{\sqrt{3} + i} = \frac{\sqrt{3} -i}{3+1} = \frac{1}{4} (\sqrt{3} - i) = \frac{1}{4} \left [2 \left \angle -\frac{\pi}{6} \right][/tex]

    You could generalise the same thing for any [itex]\mbox{a,b} \in \Re[/itex] for any complex expression a+bi.

    EDIT: To expand on this note that by De Moivre's formula:

    [tex](cos \theta + isin \theta)^n = cos(n \theta) + isin(n \theta)[/tex]

    Let [tex]z = r (cos \theta + isin\theta)[/tex].
    [tex]z^n = r^n (cos(n \theta) + i sin(n \theta))[/tex]

    Let n=-1, as in your case where we want to find the reciprocal.

    [tex]\frac{1}{z} = \frac{1}{r} (cos(-\theta) + isin(-\theta)) = \frac{1}{r} (cos(\theta) - isin(\theta))[/tex]

    Thereby the sign changes from '+' to '-'.
     
    Last edited: May 20, 2008
  6. May 20, 2008 #5

    Dick

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    1/cis(theta)=cis(-theta). Try and show that as an exercise. Hint: show cis(theta)*cis(-theta)=1 because sin(theta)^2+cos(theta)^2=1.
     
  7. Apr 13, 2011 #6
    I have a problem with a question about polar form, could you help me please?

    z=3(cos100degree+isin100degree)

    give the polar form of 1/z.
     
  8. Apr 13, 2011 #7

    Dick

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    Posting your own thread is better than tagging onto another one, but ok. What's cis(100*degree)*cis(-100*degree)? I'm assuming you know cis(a)=cos(a)+i*sin(a).
     
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