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Math Help - Trouble with complex numbers

  1. #1
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    Trouble with complex numbers

    The question is this:

    |(2+i)(cos\frac{\pi}{6}-i sin\frac{\pi}{6})|

    I'm not sure if what I am doing is right...

    My next step was this:
    |(2+i)(\frac{\sqrt3}{2}-\frac{i}{2})|

    But I'm not sure what to do after. Is it just simple algebra? Because if it is, I did something really weird and didn't get the answer, which is supposed to be \sqrt5

    Any help would be wonderful! Any working would be great, to see where I went wrong.
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  2. #2
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    Re: Trouble with complex numbers

    You're doing fine, just multiply out the brackets and write it as x+yi for some values x and y. Then in order to find the absolute value, you simply take the square root of x^2+y^2. If you picture the complex number x+yi on the complex plane, where the x axis corresponds to the real part and the y axis corresponds to the imaginary part, then |x+yi| is simply the hypotenuse of a right angle triangle with horizontal side of length x and vertical side of length y.
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  3. #3
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    Re: Trouble with complex numbers

    Quote Originally Posted by maybealways View Post
    The question is this:
    |(2+i)(cos\frac{\pi}{6}-i sin\frac{\pi}{6})|
    I'm not sure if what I am doing is right...
    My next step was this:
    |(2+i)(\frac{\sqrt3}{2}-\frac{i}{2})|
    You can really simplify things by noting |z\cdot w|=|z|\cdot |w|.
    Thus |(2+i)(\cos\frac{\pi}{6}-i \sin\frac{\pi}{6})|=|(2+i)|\cdot|(\cos\frac{\pi}{6  }-i \sin\frac{\pi}{6})|

    Moreover |\cos(\theta)\pm i \sin(\theta)|=1,~\forall\theta\in\mathbb{R}~.
    Last edited by Plato; May 7th 2012 at 02:57 PM.
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