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Math Help - theorem

  1. #1
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    theorem

    let z=r(cosθ+isinθ) be any complex number & n be any positive integer. Then zn =rn (cos(nθ)+isin(nθ)) or equivelently, [rn , nθ].

    Prove it in special cases n=2,3,4
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  2. #2
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    Re: theorem

    Quote Originally Posted by franios View Post
    Prove it in special cases n=2
    So, why don't you do this?
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  3. #3
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    Re: theorem

    I'm just confused, do I have to just plug in n? What about theta?
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  4. #4
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    Re: theorem

    Do you know how to prove that for n = 2 it is the case that for all x and y, (x + y)^n = x^n + 2xy + y^n?

    Now prove that for n = 2 it is the case that for all r and θ, (r(cosθ+isinθ))^n = rn(cos(nθ)+isin(nθ)).
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  5. #5
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    Re: theorem

    Quote Originally Posted by franios View Post
    I'm just confused, do I have to just plug in n? What about theta?
    \begin{align*}z\cdot z &=r(cos(t)+i\sin(t))r(cos(t)+i\sin(t))\\&=r^2(\cos  ^2(t)-\sin^2(t))+i(\cos(t)\sin(t))\\&=r^2(cos(2t)+i\sin(  2t))\end{align*}

    Now z^3=z^2\cdot z
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