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Math Help - Finding an exact expression for tan(pi/12)

  1. #1
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    Finding an exact expression for tan(pi/12)

    The question:
    Let z = -2 + 2i and w = -1 - \sqrt{3}i.
    i) Write z and w in polar form and thus write zw in polar form.
    ii) Hence find an exact expression for tan(\frac{\pi}{12})

    My attempt:
    i) This is simple,  z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}
    ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

    Any help would be great!
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  2. #2
    Pim
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    If you visualize this, you get a triangle, one point on the origin. The vector zw makes an angle of \frac{1}{12} \pi with the x-axis. Therefore, if you get an a+bi form of 2\sqrt{8}e^{\frac{1}{12 \pi}} you have tan(\frac{1}{12 \pi}) = \frac{b}{a}
    You can get the a+bi form by just multiplying z and w and working out the brackets.

    I hope this is clear, if not, please say so.
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  3. #3
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    Quote Originally Posted by Glitch View Post
    The question:
    Let z = -2 + 2i and w = -1 - \sqrt{3}i.
    i) Write z and w in polar form and thus write zw in polar form.
    ii) Hence find an exact expression for tan(\frac{\pi}{12})

    My attempt:
    i) This is simple,  z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}
    ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

    Any help would be great!
    \displaystyle\ tan\left(\frac{\pi}{12}\right)=\frac{sin\left(\fra  c{\pi}{12}\right)}{cos\left(\frac{\pi}{12}\right)}

    e^{(i\theta)}=cos(\theta)+isin(\theta)

    so if you multiply out

    zw=(-2+2i)(-1-\sqrt{3}i) you will be able to continue by comparing terms
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  4. #4
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    Aha, that's interesting. Thanks!
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  5. #5
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    Quote Originally Posted by Glitch View Post
    The question:
    Let z = -2 + 2i and w = -1 - \sqrt{3}i.
    i) Write z and w in polar form and thus write zw in polar form.
    ii) Hence find an exact expression for tan(\frac{\pi}{12})

    My attempt:
    i) This is simple,  z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}
    ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

    Any help would be great!
    You don't need to go into the Complex number system here.

    Just use the angle sum identity for tangent...

    \tan{(\alpha \pm \beta)} = \frac{\tan{\alpha} \pm \tan{\beta}}{1 \mp \tan{\alpha}\tan{\beta}}.


    Here \tan{\frac{\pi}{12}} = \tan{\left(\frac{\pi}{3} - \frac{\pi}{4}\right)}

     = \frac{\tan{\frac{\pi}{3}} - \tan{\frac{\pi}{4}}}{1 + \tan{\frac{\pi}{3}}\tan{\frac{\pi}{4}}}
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  6. #6
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    However, the question really is specifically asking you to go the route of the polar form of the complex numbers.
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