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Math Help - Principal arguements

  1. #1
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    Principal arguements

    I've got to find the modulus and principle argument and I'm a little unsure how to do it for this example:

    cos(pi/7) + isin(pi/7)

    I can do it perfectly on complex numbers, but I'm unsure how to start this one?

    Thanks :]
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  2. #2
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    Quote Originally Posted by scofield131 View Post
    I've got to find the modulus and principle argument and I'm a little unsure how to do it for this example:

    cos(pi/7) + isin(pi/7)

    I can do it perfectly on complex numbers, but I'm unsure how to start this one?

    Thanks :]

    What do you mean by "I can do it perfectly on complex numbers, but I'm unsure how to start this one"?? You're given a complex number!

    z=\cos \pi/7+i\sin \pi/7\Longrightarrow Re(z)=\cos \pi/7\,,\,\,Im(z)=\sin \pi/7 \Longrightarrow |z| = \sqrt{Re(z)^2+Im(z)^2}=1\,,\,\,\arg(z)=\arctan Im(z)/Re(z)=\pi/7 ....

    Tonio
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  3. #3
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    Quote Originally Posted by scofield131 View Post
    I've got to find the modulus and principle argument and I'm a little unsure how to do it for this example:

    cos(pi/7) + isin(pi/7)

    I can do it perfectly on complex numbers, but I'm unsure how to start this one?

    Thanks :]
    I should read through the entire thread before I respond! Tonio answer this half an hour ago and in almost the same words! (For some reason, this thread is still listed as having "0 responses".)

    I am confused. If you cam "do it perfectly on complex numbers" why can't you do it for this complex number?

    Do you mean you can find the modulus and principle argument for something like a+ bi? I presume, then, that you know the modulus is \sqrt{a^2+ b^2} and the "principle argument" is the principle value for the arctangent function, arctan(\frac{b}{a}).

    Then how could you have a problem here? With a= cos(\pi/7) and b= sin(\pi/7) you should be able to see immediately that \sqrt{a^2+ b^2}= \sqrt{cos^2(\pi/7)+ sin^2(\pi/7)}= 1 because sin^2(\theta)+ cos^2(\theta)= 1 for all \theta.

    And surely you can do arctan(\frac{b}{a})= arctan(\frac{sin(\pi/7)}{cos(\pi/7)})= arctan(tan(\pi/7)).

    I think the difficulty is that this problem is too easy!
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