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Thread: Trouble with complex numbers

  1. #1
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    Trouble with complex numbers

    The question is this:

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

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

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

    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 $\displaystyle \sqrt5$

    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 $\displaystyle x+yi$ for some values x and y. Then in order to find the absolute value, you simply take the square root of $\displaystyle x^2+y^2$. If you picture the complex number $\displaystyle x+yi$ on the complex plane, where the x axis corresponds to the real part and the y axis corresponds to the imaginary part, then $\displaystyle |x+yi|$ is simply the hypotenuse of a right angle triangle with horizontal side of length $\displaystyle x$ and vertical side of length $\displaystyle y$.
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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:
    $\displaystyle |(2+i)(cos\frac{\pi}{6}-i sin\frac{\pi}{6})|$
    I'm not sure if what I am doing is right...
    My next step was this:
    $\displaystyle |(2+i)(\frac{\sqrt3}{2}-\frac{i}{2})|$
    You can really simplify things by noting $\displaystyle |z\cdot w|=|z|\cdot |w|.$
    Thus $\displaystyle |(2+i)(\cos\frac{\pi}{6}-i \sin\frac{\pi}{6})|=|(2+i)|\cdot|(\cos\frac{\pi}{6 }-i \sin\frac{\pi}{6})|$

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