Results 1 to 10 of 10

Math Help - complex number and trigo Urgently

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    6

    complex number and trigo Urgently

    given that z =cosθ +jsinθ
    ((zē-1)/(zē+1))=jtanθ
    Can someone prove it to mi pls. THX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Aug 2008
    Posts
    6
    can any1 pls help mi????
    i nid help
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by juzo View Post
    given that z =cosθ +jsinθ
    ((zē-1)/(zē+1))=jtanθ
    Can someone prove it to mi pls. THX
    Substitute for z in numerator and denominator and expand.

    Substitute \cos^2 \theta - 1 = - \sin^2 \theta and 1 - \sin^2 \theta = \cos^2 \theta.

    Simplify:


    \frac{- \sin^2 \theta + j \cos \theta \sin \theta}{\cos^2 \theta + j \cos \theta \sin \theta} = \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}


    = \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}


    = \frac{j \sin \theta \left(- \frac{\sin \theta}{j} + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)}


    = \frac{j \sin \theta \left(j \sin \theta + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)} = \frac{j \sin \theta}{\cos \theta}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2008
    Posts
    6
    Quote Originally Posted by mr fantastic View Post
    Substitute for z in numerator and denominator and expand.

    Substitute \cos^2 \theta - 1 = - \sin^2 \theta and 1 - \sin^2 \theta = \cos^2 \theta.

    Simplify:


    \frac{- \sin^2 \theta + j \cos \theta \sin \theta}{\cos^2 \theta + j \cos \theta \sin \theta} = \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}


    = \frac{\sin \theta (- \sin \theta + j \cos \theta)}{\cos \theta (\cos \theta + j \sin \theta)}


    = \frac{j \sin \theta \left(- \frac{\sin \theta}{j} + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)}


    = \frac{j \sin \theta \left(j \sin \theta + \cos \theta\right)}{\cos \theta (\cos \theta + j \sin \theta)} = \frac{j \sin \theta}{\cos \theta}
    Sry, i dun get the first line. can u show a bit more of the steps of the first part
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by juzo View Post
    Sry, i dun get the first line. can u show a bit more of the steps of the first part
    Did you substitute z = \cos \theta + j \sin \theta ?

    Did you expand?

    Did you substitute and ?

    Please show your working and where you get stuck.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2008
    Posts
    6
    Given that z=cosθ+jsinθ show that
    ((cos2θ+sin2θ-1)/(cos2θ+sin2θ+1))
    den i dunno sub wht inside to get correct
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, juzo!

    I have a looong proof . . . hope someone finds a shorter one!


    Given: . z\: =\:\cos\theta +j\sin\theta

    Prove: . \frac{z^2-1}{z^2+1} \:=\:j\tan\theta

    Using DeMoivre's Theorem, we have:

    . . z^2 - 1 \;=\;(\cos\theta + j\sin\theta)^2 - 1 \;=\;\cos2\theta + j\sin2\theta - 1

    . . z^2 + 1 \;=\;(\cos\theta + j\sin\theta)^2 + 1 \;=\;\cos2\theta + j\sin2\theta - 1


    So we have: . \frac{z^2-1}{z^2+1} \;=\;\frac{(\cos2\theta-1) + j\sin2\theta}{(\cos2\theta + 1) + j\sin2\theta}


    Rationalize: . \frac{(\cos2\theta-1) + j\sin2\theta}{(\cos2\theta + 1) + j\sin2\theta} \cdot {\color{blue}\frac{(\cos2\theta+1) - j\sin2\theta}{(\cos2\theta + 1) - j\sin2\theta}}

    . . = \;\frac{\overbrace{(\cos^2\!2\theta - 1)}^{-\sin^2\!2\theta} - j\sin2\theta(\cos2\theta-1) + j\sin2\theta(\cos2\theta+1) \:\overbrace{- j^2\sin^2\!2\theta}^{+\sin^22\theta}}{(\cos2\theta  +1)^2 - j^2\sin^2\!2\theta}

    . . = \;\frac{-j\sin2\theta\cos2\theta + j\sin2\theta + j\sin2\theta\cos2\theta + j\sin2\theta}{\cos^2\!2\theta + 2\cos2\theta + 1 + \sin^2\!2\theta}

    . . = \;\frac{2j\sin2\theta}{2 + 2\cos2\theta} \;=\;\frac{j\sin2\theta}{1 + \cos2\theta} \;=\;\frac{j\cdot2\sin\theta\cos\theta}{2\cos^2\!\  theta}

    . . = \;\frac{j\sin\theta}{\cos\theta} \;=\;j\tan\theta

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by juzo View Post
    Given that z=cosθ+jsinθ show that
    ((cos2θ+sin2θ-1)/(cos2θ+sin2θ+1))
    den i dunno sub wht inside to get correct
    I was expecting to see z^2 = (\cos \theta + j \sin \theta)^2 = \cos^2 \theta + 2 j \cos \theta \sin \theta - \sin^2 \theta


    By the way, please try to use proper English, NOT den, dunno, wht, mi pls etc. (which I personally find very annoying).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Let z=\cos \theta + i \sin \theta this means z = e^{i\theta}.

    Thus, \frac{z^2-1}{z^2+1} = \frac{e^{2i\theta}-1}{e^{2i\theta}+1} = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = \frac{\sinh (i\theta)}{\cosh (i\theta)} = \frac{i\sin \theta}{\cos \theta} = i\tan \theta
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Quote Originally Posted by ThePerfectHacker View Post
    Let z=\cos \theta + i \sin \theta this means z = e^{i\theta}.

    Thus, \frac{z^2-1}{z^2+1} = \frac{e^{2i\theta}-1}{e^{2i\theta}+1} = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = \frac{\sinh (i\theta)}{\cosh (i\theta)} = \frac{i\sin \theta}{\cos \theta} = i\tan \theta
    Good post, but it's likely that juzo won't have done complex analysis to this level.

    Like Mr Fantastic says, substitute z = \cos(\theta) + j\sin(\theta) into the equation.

    So \frac{z^2-1}{z^2+1} = \frac{(\cos(\theta) + j\sin(\theta))^2 - 1}{(\cos(\theta) + j\sin(\theta))^2 + 1} = \frac{\cos^2(\theta) + 2j\cos(\theta)\sin(\theta) - \sin^2(\theta) - 1}{\cos^2(\theta) + 2j\cos(\theta)\sin(\theta) - \sin^2(\theta) + 1}.

    Then the rest of Mr Fantastic's work follows.

    Hope that helped.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. complex trigo
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 7th 2011, 04:58 AM
  2. Some complex trigo equations
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: December 17th 2010, 12:39 PM
  3. Replies: 1
    Last Post: October 2nd 2010, 01:54 PM
  4. Replies: 3
    Last Post: September 13th 2010, 11:13 AM
  5. 1^(1/3) a complex number?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2009, 10:01 AM

Search Tags


/mathhelpforum @mathhelpforum