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Math Help - Imaginary numbers with trig

  1. #1
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    Question Imaginary numbers with trig

    Hi, I wasn't too sure what category to put this in, so i'm sorry if pre-calculus is wrong.

    I came across this question but can't seem to prove it:

    If z=Cos\theta + \iota Sin\theta, show that:

    (i) |1+z|=2Cos\frac{\theta}{2}, for 0^{\circ} \leq \theta \leq 180^{\circ}

    (ii) \frac{2}{1+z}=1-\iota tan\frac{\theta}{2}, z\neq -1


    My workings:
    (i)
    |1+z|

    Cos0+\iota Sin0+Cos\theta +\iota Sin\theta

    2Cos\frac{\theta}{2}Cos\frac{-\theta}{2}+2\iota Sin\frac{\theta}{2}Cos\frac{\theta}{2}

    Cos^2\frac{\theta}{2}+\iota Sin\frac{\theta}{2}Cos\frac{\theta}{2}=Cos\frac{\t  heta}{2}

    After this I couldn't find a way to prove it and I just kept going round in circles.


    (ii)
    \frac{2}{1+z}

    \frac{1}{Cos\frac{\theta}{2}} = 1-\iota tan\frac{\theta}{2} ... from part one

    Sec^2\frac{\theta}{2} = 1-tan^2\frac{\theta}{2}-2\iota tan\frac{\theta}{2}

    Sec^2A = 1-(Sec^2A-1)-2\iota tanA

    Sec^2A = 1-\iota tanA

    tan^2A = -\iota tanA

    tanA = -\iota

    Which I don't think's possible.
    Any help here would be appreciated.
    Thanks.
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  2. #2
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    Quote Originally Posted by FreeT View Post
    Hi, I wasn't too sure what category to put this in, so i'm sorry if pre-calculus is wrong.

    I came across this question but can't seem to prove it:

    If z=Cos\theta + \iota Sin\theta, show that:

    (i) |1+z|=2Cos\frac{\theta}{2}, for 0^{\circ} \leq \theta \leq 180^{\circ}

    (ii) \frac{2}{1+z}=1-\iota tan\frac{\theta}{2}, z\neq -1


    My workings:
    (i)
    |1+z| Don't put the absolute value signs around the 1+z. At this point, you are finding an expression for 1+z, not |1+z|.

    Cos0+\iota Sin0+Cos\theta +\iota Sin\theta

    2Cos\frac{\theta}{2}Cos\frac{-\theta}{2}+2\iota Sin\frac{\theta}{2}Cos\frac{\theta}{2}
    That's correct up to that point. Also, \cos(-x) = \cos x, so you can write that last line as 1+z = 2\cos^2\tfrac\theta2 + 2i\sin\tfrac\theta2\cos\tfrac\theta2 = 2\cos\tfrac\theta2(\cos\tfrac\theta2 + i\sin\tfrac\theta2).

    Now remember that the absolute value of a complex number is given by |x+iy| = \sqrt{x^2+y^2}. Therefore |1+z| = \sqrt{(2\cos\tfrac\theta2)^2(\cos^2\tfrac\theta2 + \sin^2\tfrac\theta2)}<br />
.
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  3. #3
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    Thumbs up

    Thanks. I got it out.
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  4. #4
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    Quote Originally Posted by FreeT View Post
    (ii) \frac{2}{1+z}=1-\iota tan\frac{\theta}{2}, z\neq -1

    (ii)
    \frac{2}{1+z}

    \frac{1}{Cos\frac{\theta}{2}} = 1-\iota tan\frac{\theta}{2} ... from part one

    Sec^2\frac{\theta}{2} = 1-tan^2\frac{\theta}{2}-2\iota tan\frac{\theta}{2}

    Sec^2A = 1-(Sec^2A-1)-2\iota tanA

    Sec^2A = 1-\iota tanA

    tan^2A = -\iota tanA

    tanA = -\iota

    Which I don't think's possible.
    Any help here would be appreciated.
    Thanks.
    \frac{2}{1 + z} = \frac{2}{1 + \cos{\theta} + i\sin{\theta}}

     = \frac{2(1 + \cos{\theta} - i\sin{\theta})}{(1 + \cos{\theta} + i\sin{\theta})(1 + \cos{\theta} - i\sin{\theta})}

     = \frac{2(1 + \cos{\theta} - i\sin{\theta})}{(1 + \cos{\theta})^2 + \sin^2{\theta}}

     = \frac{2 + 2\cos{\theta} - 2i\sin{\theta}}{1 + 2\cos{\theta} + \cos^2{\theta} + \sin^2{\theta}}

     = \frac{2 + 2\cos{\theta} - 2i\sin{\theta}}{1 + 2\cos{\theta} + 1}

     = \frac{2 + 2\cos{\theta} - 2i\sin{\theta}}{2 + 2\cos{\theta}}

     = 1 - \frac{2i\sin{\theta}}{2 + 2\cos{\theta}}

     = 1 - \frac{i\sin{\theta}}{1 + \cos{\theta}}

     = 1 - i\tan{\frac{\theta}{2}}.
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