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A little help with this complex number question

  1. Oct 9, 2006 #1
    if z = 2 r cos x + r i sin x
    what is the value of lzl

    I worked for 3 hours but yet can only find lzl in terms of r and x, but the question says find the value, can anyone help solve? this is a special question to me because i always see polar forms with coefficient of the sin and cos as the same, but this question shows otherwise.

    ty, help appreciated.
     
    Last edited: Oct 9, 2006
  2. jcsd
  3. Oct 9, 2006 #2
    you sure you dont mean [tex] z = 2rCos(x) + riSin(x) [/tex] ? If so use your Cis rule or exponential equality for cos + i sin
     
  4. Oct 9, 2006 #3
    sorry bout the i, i edited already, what do u mean by the cis rule or exponential equality, can u teach me?
     
  5. Oct 9, 2006 #4
    can anybody help solve this for me?
     
  6. Oct 9, 2006 #5

    HallsofIvy

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    Am I correct, that you are given z= 2r cos(x)+ i r sin(x) and want to find |z|? That will depend on r and x- it's obviously not a single number since the larger r is obviously the larger the |z|. If, for example, x= 0 then z= 2r which has absolute value 2r. If x= [itex]\frac{\pi}{2}[/itex] then z= i r and so |z|= r.

    [tex]|z|= \sqrt{z\cdot\overlinez}= \sqrt{(2r cos(x)+ i r sin(x))(2r cos(x)- i r sin(x)}= \sqrt{4r^2cos^2(x)- r^2 sin^2(x)}[/= r \sqrt{4 cos^2(x)- sin^2(x)}[/tex]

    You might be able to simplify that squareroot by trig identities but you can't get rid of r and x.
     
  7. Oct 9, 2006 #6
    the cis thing is basically converting complex cartesian forms to polar ones

    [tex] {e^{it}} = Cos(t) + i Sin(t) [/tex]

    the t should be the power aswell with i but for some reason the tex wont do it :( need to learn more about it i guess
     
  8. Oct 9, 2006 #7
    ok, so now i know i was right, time to get to school and prove to my teacher that there is no answer, since the question asks for a value, ty ppl for helping me, i wasted 5 hours of my life working on something i got right 5 hours ago, once again ty ppl, help appreciated.
     
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